IMAGE EVALUATION TEST TARGET (MT-S) / «. <6 1.0 I.I If IIM IIIIIM " If IS 1^ 12.0 lU 11:25 i 1.4 1.6 lU-x c .^Sciences Corporation «^' [V ^\ ii^ -T^ ^^^.. <>.\ 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 '<" CP. H PREFACi:. As is evident from what lias been said, the whole intention in preparing the work has been to furnish the student with that kind of geometric knowledge which may enable him to take up most successfully the modern works on Analytical Geometry. N. F. D Queen's College, Kingston, Canada. CONTENTS. I'AGE PART I. Section I.-The Line and Point. Section II.-Two Lines- Angles. Section IIL-Three or more Lines ar:i Determined Points-The Triangle. section I V.-Parallels. Section V.-The Circle. Section VL-Constructive Geometry, . . . , PART IL Section L^Comparison of Areas. Section IL- Measurement of Lengths and Areas. Section HL-Geometric Interpretation of Algebraic Forms. Section I V.-Areal Relations-Squares and Rect- angles. Section V.-Constructive Geometry, 91 PART IIL Section I. - Proportion amongst Line - Segments. SECTION II.-Functions of Angles and their Applications in Geometry, ix 147 . ii PAGK ^ CONTENTS. PART IV. Section I. —Geometric Extensions. Section II. Centre of Mean Position. Section III.— Col- linearity and Concurrence. Section IV.— Inver- sion and Inverse Figures. Section V.— Pole aid Polar. Section VI.— The Radical Axis. Sec- tion VII.— Centres and Axes of Perspective or Similitude, ,7g PART V. > Section I.— Anharmonic Division. Section II.— Harmonic Ratio. Section III.— Anharmonic Properties. Section IV.— Polar Reciprocals and Reciprocation. Section V.— Homography and Involution, 2^3 ELEMENTARY SYNTHETIC GEOMETRY. I WOl / att£ tecl S exp: mat A seqii cour In thin^ tion purp theoi whos Ex is a t 3"- true \ PART I. GENERAL CONSIDERATIONS. 1°^ A statement which explains the sense in which some word or phrase is employed is a definition. A definition may select some one meaning out of several t^Lti:::r\ ""'' °^ '^ -^^^ ^-^rodut so" tecnnical term to be used m a particular sense. Some terms, such as space, straight, direction, etc which express elementary ideas cannot be defined. m^hem^i^atreS^^ ' ^'^ '^^"^^^ ^^^^^^^ ^' -- A theorem may be stated for the purpose of being sub- equently proved, or it may be deduced from some prev ous course of reasoning. previous In the former case it is called a Proposition, that is some tio^orrrh'' '"' ""f^ °^<"> '"^ ^'^'--' "' --^a- purpose o .!"'"' '"" ^'^ '"' ^■■^""^"' "^ P™"^- The purpose of the argument is to show that the truth of the Whose truth has already been established or admitted. is a theor?m' "" °' '"" "'^ """"""= '^ ^ ^^ ""-"er" A r a sphere, That absolute sameness (14°) which characterizes every part of a line leads directly to the following conclusions :— (i) No distinction can be made between any two segments of the same hue equal in length, except that of position in the line. (2) A line cannot return into, or cross itself. (3) A line is not necessarily limited in length, and hence, in imagination, we may follow a line as far as we please without coming to any necessary termination. This property is conveniently expressed by saying that a line extends to hijinity. 3.— The hypothetical end-points of any indefinite line are said to be points at infinity. All other points are finite points. iometry, or Geometry. said to be to enable ; and the y said to netric ele- ! does not for future 1 point is n line, an geometric 'e line, my refer- gmentj or 22°. Notation. A point is denoted by a single letter where- ever practicable, as " the point A." An indefinite line is also denoted by a single letter as "the line L," but in this case the letter l has no reference to any point. a b — ' A segment is denoted by naming its end points, as the "segment AB," where A and B are the end points. This is a biliteral^ or two-letter notation. A segment is also denoted by a single letter, when the limits of its length are supposed to be known, as the " seg- ment rt." This is a uniliteral, or one-letter notation. The term "segment" involves the notion of some finite length. When length is not under consideration, the term "line" is preferred. Thus the "line AB " is the indefinite line having A and B as two points upon it. But the "segment AB" is that portion of the line which lies between A and B. 23°. In dealing with a line-segment, we frequently have to consider other portions of the indefinite line of which the segment is a part. }t£] 10 SYNTHETIC GEOMETRY. :ir As an example, let it be required to divide the segment AB —^ ^— I ^^ into two parts whereof one shall be twice as long as the other. I o do this we put C in such a position that it may be twice as far from one of the end-points of the se-ment, A sav, as it is from the other, li. But on the indefinite line through A and B we may place C so as to be twice as far from A as from B. So that we have two points, C and C, both satisfying the condition of being twice as for from A as from B. Evidently, the point C doe^ not divide the segment AB in the sense commonly attached to the word ^/v^Wc. But on account of the similar lelat'ons held by C and C to the end- points of the segment, it is convenient and advantageous to consider both points as dividing the segment AB. When thus considered, C is said to divide the segment tntcrnally and C to divide it externally in the same manner. 24°. ^;i-/<7;;/.— Through a given point only one line can pass in a given direction. Let A be the given point, and let the segment AP mark A % ^"^^ S'ven direction. Then, of all the lines that can pass through the point A, only one can have the direction AP, and this one must lie along and coincide with AP so as to form with it virtually but one line. Cor. I. A finite point and a direction determine one line. Cor. 2. Two given finite points determine one line. For, if A and P be the points, the direction AP is given, and hence the line through A and having the direction AP is given. Cor. 1 Two lir.es by their intersection determine one finite point, ror, if they determined two, they would each pass through the same two points, which, from Cor. 2, is impossible. Cor. 4. Another statement of Cor. 2 is— Two lines which have two points in common coincide and form virtually but one line. liM •■ THE LINE AND POINT. II 25'. Axiom,~k straight line is the shortest distance be- tween two given points. Altlioucrh it is possible to give a reasonable proof of this axiom, no amount of proof could make its truth more apparent. The following will illustrate the axiom. Assume any two pomts on a thread taken as a physical line. By separating these as far as possible, the thread takes the form which we call straight, or tends to take that form. Therefore a straight finite hne has its end-points further apart than a curved line of equal length. Or, a less length of line will reach from one given point to another when the line is straight than when it is curved. A/— The distance between two points is the length of the segment which connects them or has them as end-points. 26°. Superposition.-Comparison of Figures. -We assume that space is homogeneous, or that all its parts are alike, so that the properties of a geometric figure are independent of its position in space. And hence we assume that a figure may be supposed to be moved from place to place, and to be turned around or over in any way without undergoing anv change whatever in its form or properties, or in the relations existing between its several parts. The imaginary placing of one figure upon another so as to compare the two is called superpositiotu By superposi- tion we are enabled to compare figures as to their equality or inequality. If one figure can be superimposed upon another so as to coincide with the latter in every part, the two figures are necessarily and identically equal, and become virt'Iially one figure by the superposition. 2f. Two line-segments can be compared with respect to length only. Hence a line is called a magnitude of one dimension. Two segments are ^$r«rt/ when the end-points of one can be i Iff! 12 SYNTHETIC GEOMETRY. !i M III made to coincide with the end-points of the other by super- position. 28°. Z>^— The sum of two segments is that segment which is equal to the two when placed in line with one end-point in each coincident. Let AB and DE be two segments, and on the line of which 9^ f AB is a segment let BC be equal to _^ ^ ^ DE. Then AC is the sum of AB A B c and DE. This is expressed symbolically by writing AC=AB-f-DE, where = denotes equality in length, and -I- denotes the placing of the segments AB and DE in line so as to have one common point as an end-point for each. The interpretation of the whole is, that AC is equal m length to AB and DE together. 29°. Z><^— The difference between two segments is the segment which remains when, from the longer of the segments, a part is taken away equal in length to the shorter. Thus, if AC and DE be two segments of which AC is the longer, and if BC is equal to DE, then AB is the difference between AC and DE. This is expressed symbolically by writing AB=AC-DE, which is interpreted as meaning that the segment AB is shorter than AC by the segment DE. Now this is equivalent to saying that AC is longer than AB by the segment DE, or that AC is equal to the sum of AB and DE. Hence when we have AB=AC- DE we can write AC = AB -i- DE. We thus see that in using these algebraic symbols, ==,-!-, and -, a term, as DE, may be transferred from one side of THE LINE AND POINT. 13 the equation to another by changing its sign from + to - or vice versa. Owing to the readiness with which these symbolic expres- sions can b«.' manipulated, they seem to represent simple alge- braic relations, hence beginners are apt to think that the work- ing rules of algebra must apply to them as a matter of necessity. It must be remembered, however, that the formal rules of algebra are founded upon the properties of numbers, and that we should not assume, without examination, that these rules apply without modification to that which is not number. This subject will be discussed in Part II. 30°. Z>^.— That point, in a line-segment, which is equi- distant from the end-points is the middle point ^^f the segment. It is also called the internal point of bisection of the seg- ment, or, when spoken of alone, simply ihepoijit 0/ bisection. Exercises. I. 2. If two segments be in line and have one common end- point, by what name will you call the distance between their other end-points } Obtain any relation between " the sum and the differ- ence " of two segments and "the relative directions "of the two segments, they being in line. 3. A given line-segment has but one middle point. 4. In Art. 23°, if C becomes the middle point of AB, what becomes of C 1 5. In Art. 30° the internal point of bisection is spoken of. What meaning can you give to the "external point of bisection " } m\ fa s^ •:i l!'- i'l '"il m Iff' 14 SYNTHETIC GEOMETRY. SECTION II. RELATIONS OF TWO LINES.— ANGLES. 31°. When two lines have not the same direction they are said to make an a»or/e with one another, and an aug/e is a c difference in direction. Illiistration.— 'L^X. A and B represent two stars, and E the A' »A position of an observer's eye. Since the lines EA and EB, which join the eye and the stars, have not the same direction they make an angle with one another at E. I. If the stars appear to recede from one another, the angle at E becomes greater. Thus, if B moves into the position of C, the angle between EA and EC is greater than the angle between EA and EB. Smiilarly, if the stars appear to approach one another, the angle at E becomes smaller; and if the stars become coinci- dent, or situated in the same line through E, the angle at E vanishes. Hence an angle is capable of continuous increase or dmimution, and is therefore a magnitude. And, being magnitudes, angles are capable of being compared with one another as to greatness, and hence, of being measured. 2. If B is moved to B', any point on EB, and A to A', any point on EA, the angle at E is not changed. Hence increas- ing or diminishing one or both of the segments which form an angle does not affect the magnitude of the angle. Hence, also, there is no community in kind between an angle and a line-segment or a line. Hence, also, an angle cannot be measured bv means of line-segments or lines. RELATIONS OF TWO LINES.— ANGLES. 1 5 32°. Def.—k line which changes its direction in a plane while passing through a fixed point in the plane is said to rotate about the point. The point about which the rotation takes place is the polc^ and any segment of the rotating line, having the pole as an end-point, is a radius vector. Let an inextensible thread fixed at O ^p ' be kept stretched by a pencil at P. Then, when P moves, keeping the thread straight, OP becomes a radius vector rotating about the pole O. When the vector rotates from direction OP to direction OP' it describes the angle between OP and OP'. Hence we have the following : — Def. I. — The ajif^le between two lines is the rotation neces- sary to bring one of the lines into the direction of the other. The word "rotation," as employed in this definition, means the amount of turning effected, and not the process jf turning. Def 2.— For convenience the lines OP and OP', which, by their difference in direction form the angle, are called the arms of the angle, and the point O where the arms meet is the vertex. Cor. From 31°, 2, an angle does not in any way depend upon the lengths of its arms, but only upon their relative directions. ^L 33°. Notation of Angles.— \. The symbol l is used for the word "angle." 2. When two segments meet at a vertex the angle between them may be denoted by a single letter ^--^^ placed at the vertex, as the ^O , or by a letter with or without an arch of dots, ° as ^ ; or by three letters of which the extreme ones denote points upon the arms of the angle and the middle one denotes the vertex, as ^OB. ^Wf: i6 SYNTH i:tic geometry. liiU mi 3. The angle between two lines, when the vertex is not pictured, or not referred to, is expressed by z.(L. M), or LM (t> o^ln PT:'' """ ^" ^^^ one-letter\o:^i!!; t Id^^^^-'-' ^^ -' ^^ ^-- ^^e lines in r>cA~Two angles are equal when the arms of the one o may be made to coincide in direction respec- tively with the arms of the other ; or when the angles are described by the same rotation. Thus, If, when a is placed upon O, and O A is made to lie along OA, O'B' can also B' be made to lie along OB, the ^A'O B' is equal to lAOB. This equality is symbolized thus • ^A'0'B'=zj\OB. -A' Where the sign = is to be interpreted as indicating the possibility of coincidence by superposition Jl' ^-''"u """"^ J^^ffercnce of An^/es.~The sum of two angles IS the angle described by a radius vector whilh describes the two angles, or their equals, in succession. P ' Thus if a radius vector starts from co- incidence with OA and rotates into direction OP it describes the ^AOP. _ If It next rotates into direction OP' it A describes the /.POP' Rnf in ;f. u i rotation it has described the /.AOpt Th;refore ^A0P'=aA0P + ^P0P' Similarly, z.AOP=^AOP'-^POP'. . ^'^/-When two angles, as AOP and POP' have one arm xr::if '^"^^" ''- '--'^-^ - ' *« -^•-™ 36". De/.-A radius vector which starts from any given d,rect,on and makes a complete rotation so as to return f^fts or,g,nal d.rect.on describes a »numa«^;e, or perigon I ' 1 : RELATIONS OF TWO LINES.-ANGLES. j; One-half of a circumangle is a straight angle and one fourth of a circumangle is a right angle. .1,^^°" ^'^'f'f^t"^^ ''"^ ""'^^^' °^ ^^"es meet in a point OrOB OC tr" r"^^^^^ '°^"^^' ^^ ^ circumanX' UA, UB, OC, ..., OF are hnes meetino- in O. Then ^ ^0B+z.B0C + lC0D + ...+^F0A = a circumangle. D. Proo/.~K radius vector which starts from comcidence with OA and rotates into the successive directions, OB, OC E0F?F0a!'"'''' " '"'''''^°'' '^^ ^'"^les AOB, BOC,""..., But in its complete rotation it describes a circumangle (36°) •• ^OB + z.BOC+...+^FOA=a circumangle l.i* Lor. The result may be thus stated •— The sum of all the adjacent angles about a point in the plane IS a circumangle. side'lVrtr'T"'!;' '"" °' '" ^'^ ^'J'^^^"^ ^"gl^^ on one side of a Ime, and about a point in the line IS a straight angle. O is a point in the line AB ; then ^OC + ^COB=:a Straight angle, b ^ Proo/.~Le\i A and B be any two points m the Ime, and let the figure formed by \c- nl^A ^^ t^ '^''^^^"^ "^^"^ ^B ^v'thout displacing the Then (24 , Cor. 2) O ,s not displaced by the revolution, Z-AOC=^AOC', and ^BOC = ^BOC'- ••• ^OC-f-^BOC=^AOC'+/.BOC' ' therefore the sum of each pair is a straight angle (36°). .la is a ::;4i!S'' '''"'" ^'^ '^'""^ '"^^^'^"^ °^ ^ ^- B !!! i8 SYNTHETIC GEOMETRY. iii 11 H ■ ■ i.,j ■ ' ' ': i 1 1- 1 ■ f - Cor. 2. If a radius vector be rotated until its direction is reversed it describes a straight angle. And conversely, if a radius vector describes a straight angle its original direction is reversed. Thus, if OA rotates through a straight angle it comes into the direction OB. And conversely, if it rotates from direction OA to direction OB it describes a straight angle. 39° When two lines L and M cut one another four angles are formed about the point of intersection, any one of which may be taken to be the angle between the lines. These four angles consist of two pairs of opposif^ or v,rrica/ angles, viz., A, A', and B, B', A being opposite A', and B being opposite B'. 40°. 77/^^7;-^;;/. -The opposite angles of a pair formed by two intersecting lines are equal to one another. Proof.— ^A + ^B = a straight angle (38°) ^A'+^B = a straight angle. ('■^8°') ^A=z.A', ^ ^ q.e.ct. Def. I.— Two angles which together make up a straio-ht angle are supplementary to one another, and one is called the supplement of the other. Thus, A is the supplement of B' and B of A'. Cor. If £A = .LB, then ^A'=^B=^B', and all four angles are equal, and each is a right angle (36°). ^ Therefore, if two adjacent angles formed by two intersect- ing lines are equal to each other, all four of the angles so formed are equal to one another, and each is a right angle. Def. 2.-When two intersecting lines form a right angle at their point of intersection, they are said to be perpendicular to one another, and each \s perpendicular to the other and and direction is versely, if a al direction comes into im direction I M cut one :d about the le of which fie between opposite or pposite A', formed by (38°) (38°) g.e.d. a straight called the lent of B', )ur angles intersect- angles so It angle. t angle at '^endicular ler. RELATIONS OF TWO LINES.-ANGLES. ,9 Perpendicularity is denoted by the symbol 1 »„ 1, •■perpe,vdicular to" or "is perpendicular" ■" ^' ° "' "^^ A nght angle is denoted by the symbol 1 Jhe symboU also denotes two right angles or a straight ^^;/.//.,«.;,/ of the other ' ^""^ ^^'^ ^^ ^^^ .:ni"o:f:n-Si\£-;:t/4-^^^^^^^ rTgltan^: ' ^"^ "^ °^ "^^^ ''^ » ^^^"-r a," ^o^ Def. 4. — An aaite an^rle is l^cc fT,o« • i ^^/«.. angle is gre.ter thnn V "^^' ^"^'^' ^"^ ^^ : right angles ^ '" " '^^^' ""^^^' ^"^ ^^^^ than two 41°. From (36°) we have I circumangle = 2 straight angles =4 right angles. i ro1Z;i:*ni.~^^''^ '' "^^^ ^« -P--" 'n by 2. ^ '^"t.ies, and a straight angle Angles less than a right ande mav T.a « ;^.,y^at ieast, by .fctionsf ^^^tir S-Ttt eiTp^ '^ier^eg^'-^^:?,^',:-' ^ !^ f '"^^ ■■""' '° equal parts called minuTes .' and each " '' '^'° ^ parts called seconds. ' * """"'" '"'° ^o equal , Thus an angle which is one-seventh of a -V- conta.ns fifty-one degrees twenty fi, « "-"-cumangle ' c^rees, twenty-five mmutes, and forty-two I) 'i I'd M ' III 20 SYNTHETIC GEOMETRY. seconds and six-sevenths of a second. This is denoted as follows : — ri° 'yr' A^r," 51 25' 42? B 42 . r//^^rm.-Through a given point in a line only one perpendicular can be drawn to the line. The line OC is ± AB, and OD is any other line through O. Then OD is notX AB. Proo/.~-The angles BOC and COA are each right angles (40°, Def 2). rhei-efore BOD is not a right angle, and OD is not ± AB. But OD is any line other than OC. Therefore OC is the only perpendicular. g,,a: /;^_The perpendicular to a line-segment through its middle point is the r/o-/a bisector of the segment. Since A segment has but one middle point (30°, Ex. 3) and since but one perpendicular can be drawn to the segment through that point, .-. a Imc-scc^ment has but one right bisector. 43°. Def.-ThQ lines which pass through the vertex of an angle and make equal angles with the arms, are the bisectors of the angle. The one which lies within the angle is the tnfernni bisector, and the one lying without is the external bisector. Let AOC be a given angle ; and E let EOF be so drawn that Z.AOE = ^EOC. EF is the internal bisector of the angle AOC. Also, let GOH be so drawn that ° Z.COG = ^HOA. HG is the external bisector of the angle AOC. ^COG=^HOA and ^HOA=^GOB, -COG=^GOB; (hyp.) (40°) is denoted as RELATIONS OF TWO LINES.—ANGLES. 21 and the external bisector of AOC is the internal bisector of Its supplementary angle, COB, and vice versa The reason for calling GH a bisector of the angle AOC is given m the definition, viz., GH makes equal angles with the arm. Also, OA and OC are only parts'of indefinite lint as the "oB "^''•^^^''°" "^^^ ^e taken as the ^OC o; totfi°'-i"l^' ^" '^° ^^ ^°""^ '^° P°^"*^ ^hich are said o d vide the segment in the same manner, so we may find two hnes dividing a given angle in the same manner, one dmdmg It internally, and the other externally FOr'U^ ^^ '^^l ^'^^'' '^^' '^" ^^OE is double the ifdou'birtT: Igoc!" "^^^ ^'° '^ ^^^^^ - ^^- ^^^ ^^^ This double relation in the division of a segment or an angle is of the highest importance in Geometry. EF and GH are bisectors of the ^AOC • then EF is J. GH. ' Jr^'~ ^^^C=^^OC, V OE is a bisector, and ^C0G=i^C0B, v OG is a bisector • adding, ^E0G = *^0B. bisector, But ^^^^i:^^f-J^ht angle, (33=, Cor. .) ^EOG is a right angle. ^^^o^ r. Exercises. Three lines pass through a common point and divide the plane mto 6 equal angles. Express the value of each angle ,„ nght angles, and in degrees. OA and OB make an angle of 30°, how many degrees are there m the antrle made hv o 1 j 1 )• ""^Jj^es are of the angle KolT ' """ ''''™"' '^'^^"°'- if '>'•> SYNTHETIC GEOMETRY. 3. What is the supplement of 13° 2/ 42"? What is its complement ? 4- Two lines make an angle a with one another, and the bisectors of the angle are drawn, and again the bisectors of the angle between these bisectors. What are the angles between these latter lines and the original ones ? 5. The lines L, M inteisect at O, and through 0, L' and M' are drawn J. respectively to L and M. The angle be- tween L' and M' is equal to that between L and M. !! , SECTION III. THREE OR MORE POINTS AND LINES. THE TRIANGLE. 46°. 77/^^;r;«.— Three points determine at most three lines ; and three lines determine at most three points. Proof I.— Since (24°, Cor. 2) two points determine one line, three points determine as many lines •^as we can form groups from three points ^i_ taken two and two. Let A, B, C be the points ; the groups are AB, BC, and CA. Therefore three points determine at most three lines. 2.— Since (24°, Cor. 3) two lines determine one point, three lines determine as many points as we can form groups from three lines taken two and two. But if L, M, N be the lines the groups are LM, MN, and NL. Therefore three lines determine by their intersections at most three points. What is its THREE OR MORE POINTS AND LINKS. 23 47°. T/u'on',u.-Four points determine at most six lines • and four Imes determine at most six points. " * y'roo/-.i. Let A, 13, C, D be the four pomts. The groups of two are AB AC AD, BCBD, and CD; or six in all. ' ' Therefore six lines at most are deter- mmed. 2. Let L, M, N, K be the lines. The groups of two that can be made are KL KM,KN,LM,LN,andMN;orsixinall' Therefore six points of intersection at - most are determined. ^ Cor. In the first case the six lines determined mss bv threes through the four points. And in the second ca"e the S.X po,nts determined lie by threes upon the four lines hi^er^Vomr "' 1 '"'"'^ " ^"'^ ^"^^^^'^^ -^ in the ni^ner Geometry is of special importance. Ex Show that 5 points determine at most 10 lines and c nes determme at most 10 points. And that in th first case hi pltsTb ^'^"'^ '^^■°"^^' ^^^^ p°'"^ ■' -^ "-hel t^ me pomts lie by fours on each line. nnfl^^~'^/''''^''^'^' '' ^^'^ ^-"'■^ ^^'•"^ed by three lines mined linr^"^' '''''''' °^ '' ^'^^^ ^^^"^^ -^ ^^^ ^eT- The points are the vcr/ices of the triangle, and the line segments which have the points as end-poinl ^re ^e .! jr SDoken 'T''"'"? ^°r'°"' °^ '^' determined lines are usually ge mn;/ ".^'^ ^'^^^ P-^"-^-" But in many cases geneiality requires us to extend the term side " to the whole line. \/f Thus, the points A, B, C are the ^ vertices of the triangle ABC. The segments AB, BC, CA are the ^ sides. The portions AE. BF, CD etc extending outwards as far as requi'red, ;re the sides produced. ii i iiti ' 'ill flf 24 SYNTHETIC GEOMKTRY. 'm ilii The trian^'Ie is distinctive in bein- the rectilinear figure for which a given number of lines determines the same number of pomts, or v/ce versa. Hence when the three points, forming the vertices, are given, or when the three lines or line-segments forming the sides are given, the triangle is com- pletely given. This is not the case with a rectilinear figure having any number of vertices other than three. If the vertices be four in number, with the restriction that each vertex is determined by the intersection of two sides, any one of the figures in the margin will satisfy the conditions. Hence the giving of the four vertices of such a figure is not sufficient to completely determine the figure. 49°. I),/.-i. The angles ABC, BCA, CAB are the m- ternal angles of the triangle, or simply the angles of the triangle. 2. The angle DCB, and others of like kind, are external angles of the triangle. 3- In relation to the external angle DCB, the angle BCA IS the adjacent internal angle, while the angles CAB and ABC are opposite internal angles. 4. Any side of a triangle may be taken as its base, and then the angles at the extremities of the base are its basal angles, and the angle opposite the base is the vertical angle. The vertex of the vertical angle is the vertex of the triangle when spoken of in relation to the base. 50°. Notatio7i.~T\iQ symbol A is commonly used for the word triangle. In certain cases, which are always readily apprehended, it denotes the area of the triancrle ire external TllKKM OR MORE POINTS AND LINES. 25 The angles of the triangle are denoted usually by the capual letters A H, C, and the sides opposite by the cor- responding small letters a, b, c. ^ .^'V^;^-^^''^^" t^^« fipr^i-es compared by superposition co.nc.de m all their parts and become virtual^^x-t'one figure they are said to be conoruc/U. ^ Congruent figures are distinguishable from one another o..ly by their position in space and are said to be identically Congruence is denoted by the algebraic symbol of identity = ; and this symbol placed between two figures capable of congruence denotes that the figures are congruent ' way "th '7'; ''' '""f '' ^'"^^ °^ ^°-P--- - two ways. The first is as to their capability of perfect coinci- dence ; when this is satisfied the figures are congruent. The s cond IS as to the magnitude or extent of the portions of The plane enclosed by the figures. Equality in this respect is expressed by saying that the figures are eguaL ^ cas?withT!fe'"' '"' «f--P-ison is possible, as is the case with hne-segments and angles, the word ajuai is used. CONGRUENCE AMONGST TRIANGLES. sidf an?r;:7TT 'T^'^' -^^^ -"^^"-^ when two sides and the included angle in the one are respectively equal to two sides and the included angle in the other ^ IfAB=A'B'| the triangles ^b ^b' BC=B'C'| are congru- and z.B=^B'J ent. Proof.— V\3.ce AABC on A c a^ -^^ AA'B'C' so that B coincides with B', and BA lies alon- B'A' ^B=z.B', BC lies along B'C', " /,,o: andv AB = A'B'andBC = B'C'; ^^^ ^ A coincides with A' and C with C, (27°) I fit t ' i J.;, . ; iilll 26 SYNTHETIC GEOMETRY. iiil' ^"^;^ A . ^C lies along A'C ; (24°, Cor. 2) and the As coinciding in all their parts are congruent. (51°) q.e.d. Cor. Since two congruent triangles can be made to coin- cide m all their parts, therefore— When two triangles have two sides and the included an-le in the one respectively equal to two sides and the includ^'ed angle in the other, all the parts in the one are respectively equal to the corresponding^ parts in the other. 53°. Theorem.~T.xQrY point upon the right bisector of a segment is equidistant from the end-points of the segment. AB is a line-segment, and P is any point on its right bisector PC. Then PA = PB. Proof.— In the As APC and BPC, AC = CB, (42°, Def.) ^ACP = ^BCP, (42°, Def.) and PC is common to both As ; AAPC-ABPC, ' (52°) PA=PB. {S2% Cor.) q.e.cl. Def. I.— A triangle which has two sides equal to one an- other is an isosceles triangle. Thn.s the triangle APB is isosceles. The side AB, which is not one of the equal sides, is called the base. Cor. I. Since the AAPC=ABPC, z.A=z.B. Hence the basal angles of an isosceles triangle are equal to one another. Cor. 2. From (52°, Cor.), ^APC=z.BPC ; Therefore the right bisector of the base of an isosceles tri- angle is the internal bisector of the vertical angle. And since these two bisectors are one and the samp Hn<- th« /^^n-— :^ true. ) one an- THREE OR MORE POINTS AND LINES. 2/ Dcf. 2.-A triangle in which all the sides are equal to one another is an equilateral triangle. Cor. 3. Since an equilateral triangle is isosceles with re- spect to each side as base, all the angles of an equilateral triangle are equal to one another ; or, an equilateral triangle is equiangular. "ttii,ic 54°. rW;//.-Every point equidistant from the end- pomts of a line-segment is on the right bisector of that segment. (Converse of 53°.) PA = PB. Then P is on the right bisector of AB. A i'7''-^~^^ P 's "°t on the right bisector of AB, let the right bisector cut AP in Q Then QA = QB, ' (53O) '"^ PA = PB, (hyp.) QP = PB-QB, o; PB = QP + QB, which IS not true. Therefore the right bisector of AB does not cut if ^ and similarly it does not cut BP ; therefore it passes hrough P or P is on the right bisector. ^nrougn i^, This form of proof should be compared with that of'Art SZ , they being the kinds indicated in 6° This latter or indirect form is known as proof by reductio ad absurdum (leading to an absurdity). In it we proCe t conclusion of the theorem to be true bv show ngXt the acceptance of any other conclusion leads us to some ret on which IS absurd or untrue. relation the mid^i^7fT^ ^'"'■^^^'^^"^ ^^^"^ ^ vertex of a triangle to the^middle of the opposite side is a median of the triangle. Every triangle has three medians. Tn, Cor. 2. The median to the base of an isosceles triangle is Kfi li 28 \ mil J t'il!"i ill I ii I W§ SYNTHETIC GEOMETRY. the right bisector of the base, and the internal bisector of the vertical angle. rr-,° r- ^ (53 , Cor. 2.) Cor. 3. The three medians of an equilateral triangle are the three right bisectors of the sides, and the three internal bisectors of the angles. 56°. T/i,arem.-l{ two angles of a triangle are equal to one another, the triangle is isosceles, and the equal sides are opposite the equal angles. (Converse of k^", Cor. I.) z.PAB=^PBA, then PA = PB. Proo/.~l{ P is on the right bisector of AB, PA = PB. (53O) If P is not on the right bisector, let AP B cut the right bisector in Q. QA=QB, and /-QAB=z.QBA. (53° and Cor. i) ^PBA = ^QAB; (^yp. z.PBA=^QBA, ^ ^^^ which is not true unless P and Q coincide. Therefore if P is not on the right bisector of AB, the Z.PAB cannot be equal to the Z.PBA. But they are equal by hypothesis ; P is on the right bisector, ""^ PA=PB. . ^.,.^. Cor. If all the angles of a triangle are equal to one another, all the sides are equal to one another. Or, an equiangular triangle is equilateral. 57°. From 53° and 56° it follows that equality amongst the sides of a triangle is accompanied by equality amongst the angles opposite these sides, and conversely. Also, that if no two sides of a triangle arc equal to one another, then no two angles are ec.ual to one another, and conversely. THREE OR MORE POINTS AND LINES. 29 Def.~K triangle which has no two sides equal to one another IS a jrrt:/^«^ triangle anther.' ' "'""' '™"^" '"' "° '"° ^"S'«^ ^-J-' '<> one 58-. Theorem -IH^o triangles have the three sides in the one respecfvely equal to the three sides in the other the triangles are congruent. _ IfA'B'=ABl B'C' = BC ca'=caJ (27°) the As ABC ■and A'B'C'are congruent. A< /'r^^/— Turn the AA'B'C over and place A' on A, and A'C along AC, and let B' fall at some point D ".'"aat.^- ^'^' = AC,C falls ate, and AADC ,s the AA'B'c in its reversed position, bince AB=ADandCB = CD, ^^l^ofBa "^ ''" ''^^'" °f ^°- -" ^^ '^ *« 'isht z.BAC=^DAC; /^.^ ri^^^l and the As BAC and DAC are congruent. ^^' ' "\'^, AABC ^AA'B'C. fj 'jf'A f ^ff 'f'^-^f two triangles have two angles and the ;"d:?suet tr :r ::ri^ '° '-' -^-^'^^ jj. '"* °"'^'^' "le triangles are congruent. and ■ Ar''=Acj """ ^' ^^"^ ''""' '^''^'^' "'" '""^ruent. B' B Proof. ~.V\^^^ K on A, and A'C along AC. I'll i III II ! \i, 30 SYNTHETIC GEOMETRY. Because and V and V A'C'=AC, C coincides with C ; lA'=lA, A' IV lies along AB ; ^C=z.C, CB' lies along CB ; B' coincides with B, and the triangles are congruent. (27°) (24°, Cor. 3) q.e..d. 60" Thcorem.~hxi external angle of a triangle is greater aB than an internal opposite angle. The external angle BCD is greater than the internal opposite angle ABC or \c D BAC. Proof.— h^i BF be a median produced until FG = BF. Then the As ABF and CGF have (construction) (55°) (40°) (52°) (52°, Cor.) (40°) , BF = FG, AF = FC, and z.BFA = ^GFC. AABF^ACGF, and z.FCG = ^BAC. But zACE is greater than ^FCG. z.ACEis>ABAC. Similarly, ^BCD is>^ABC, and ^BCD=._ACE. Therefore the z^ BCD and ACE are each greater than each of the /.s ABC and BAC. ^^^gj^ 61°. Theorem.- -Ovi\y one perpendicular can be drawn to a B line from a point not on the line. Proof.~Ltt B be the point and AD the line ; and let BC be ± to AD, and BA be anv line other than BC. ° Then ^BCD is>z.BAC, (60°) .iBAC is not a "], BA is not ± to AD. But BA is any line other than BC ; BC is the only perpendicular from B to AD. g.e.d. A c and e IS greater THREE OR MORE POINTS AND LINES. 31 Cor. Combining this result with that of 42° we have- Through a given point only one perpendicular can be drawn to a givjn line. 62° T/ieorem.-Oi any two unequal sides of a triangle and the opposite angles— 1. The greater angle is opposite the longer side 2. The longer side is opposite the greater angle." I. BAis>BC; then ^Cis>^A. Proo/.~htt BD = BC. Then the ABDC is isosceles, ''^"^^ /-BDC=^BCD. But ^BDCis>z.A, nnd ABCAis>z.BCD; ^BCAis>z.BAC; °''' /-Cis>z.A. 2. Z.C is > lA ; then AB is > EC. Proo/.~Yrom the Rule of Identity (7°), since there is but one longer side and one greater angle, and since it is shown r) that the greater angle is opposite the longer side, therefore the longer side is opposite the greater angle. g,e.d. Cor. I. In any scalene triangle the sides being unequal to one another, the greatest angle is opposite the longest side and the longest side is opposite the greatest angle. Also, the shortest side is opposite the smallest angle, and conversely. Hence if ^, B, C denote the angles, and a, b, c the sides respectively opposite, the order of magnitude of A, B, C is the same as that of a, b, c. 63°. T/ieorem.~0( all the segments between a given point and ajine not passing through the point— 1. rhe perpendicular to the line is the shortest. 2. Of any two segments the one which meets the line (53°, Cor. I) (6o') q.e.d. it>. i. I 32 SYNTHETIC GEOMETRY. M4\ I' Hill further from the perpendicular is the longer ; and con- versely, the longer meets the line further from the perpendicular than the shorter does. 3. Two, and only two segments can be equal, and they lie upon opposite sides of the perpendicular. P P is any point and BC a line not pass- ing through it, and PA is J. to BC. I. PA which is ± to BC is shorter than any segment PB which is not J. Proof.- But and .• B to BC. ^PAC=z.PAB=-l. z.PACis>^PBC; z.PABis>^PBC, PBis>PA. (hyp.) (60°} (62°, 2) q.e.d 2. AC is > AB, then also PC is > PB. P;7?^/ -Since AC is > AB, let D be the point in AC so that AD-AB. Then A is the middle point of BD, and PA is the right bisector of BD. (42° Def ) PD = PB ' (53°) and z.PDB = ^PBD. (53°, Cor i) But ^PDBis>^PCB; .lPBDis>^PCB, and PCis>PB. (62°, 2) The converse follows from the Rule of Identity. q.e.d. 3. Proof.— In 2 it is proved that PD = PB. Therefore two equal segments can be drawn from any point P to the line BC ; and these lie upon opposite sides of PA. No other segment can be drawn equal to PD or PB. For it must lie upon the same side of the perpendicular, PA, as one of them. If it lies further from the perpendicular than this one it is longer, (2), and if it lies nearer the perpendicular it is shorter. Therefore it must coincide with one of them and is not a third line. g^^d THREE OR MORE POINTS AND LINES. ^3 Z;,y:_The length of the perpendicular segment between any pomt and a line is the ,is^a..e of the point from the 5oint in AC 64'. 7-W« If two triangles have two angles in the one respec .vely equa to two angles in the other, and a side opposite an equal ande in pirh « 1 v rnn.n,.n. ^ ^'^'^ ^^"''^'^ ^^^ tnangles are congruent. As B. If ^A'=zj\]then the ^, .. ^C'=^C [a'B'C and ABC A and A'E' = AbJ are congruent. / V. Proo/.—Flacc A' on A, and Z j\ A'B' along AB. ^ dc A'B' = AB, B' coincides with B. Also, •.• ^A'=z.A, A'C lies along AC. , ,. , Now If C does not coincide with C, let it fall at some other point, D, on AC. Then, •.• AB = A'B', AD = A'C', and lA^lA', (271 (34°) and But (52°) (52°, Cor.) (hyp.) .-. AA'B'C = AABD, ^ADB lC. •'. ^ADB = ^C, which is not true unless D coincides with C Therefore C must fail at C, and the As ABC and A'B'C are congruent. The case in which D may be supposed to be a point on AC produced is not necessary. For we may then super- impose the AABC on the AA'B'C. 65°. T^^^orem.~lf two triangles have two sides in the one respectively equal to two sides in the other, and an angle opposite an equal side in each equal, then— 1. If the equal angles be opposite the longer of the two sides in each, the triangles are congruent. 2. If the equal angles be opposite the shorter of the two c Is 34 SVN TI r ETIC C EOM I :TR Y. A P sides in each, the trianjTlcs arc not necessarily con- B B' ^''■"' ''^■ A'ir = AI5, c' I. If HCis>AB, AA'H'C'-AABC. Proof.— Since HC is > AB, therefore li'C is > A'B'. Place A' on A and A'C alon<,' AC. ^A' = ^A, and A'B' = AB, B' coincides with B. (34°^ 27") Let BBbe±AC; then B'C cannot lie between BA and BP (63°, 2), but must lie on the same side as BC; and beinj; e(|ual to BC, the lines B'C and BC coincide (63", 3), and hence AA'B'C'-AABC. ^,eu/. ^' 2. U BC is < AB, the As A'B'C and ABC may or may not be congruent. Proo/.—S'mce AB is > BC, PA is > PC. (63", 2) Let PD = PC, then BD-BC. Now, let AA'B'C be superimposed on AABC so that A' coincides with A, B' with B, and A'C lies along AC. Then, since we are not given the length of A'C, B'C may coincide with BC, and the As A'B'C and ABC be congruent; or B'C may coincide with P)D, and the triangles A'B'C and ABC be not congruent. ^,^.^^_ Hence when two triangles have two sides in the one respectively equal to two sides in the other, and an angle opposite one of the equal sides in each equal, the triangles are not necessarily congruent unless some other relation exists between them. TIIRKK OR MORK I'OINTS ANI ) MNKS. 35 The first part of the theorem jrivcs one of the suffici- ent relations. Others arc given in the following cor- ollanes. ^ >' , and l,e A-s AI'>I) and AHC become one and tl,e san>e. Hence C must fall at C, and the As A'H'C and AIJC are congruent. f ^urA' '?^^^^^^'''"I^f^'^'"^"tary to BDC and therefore to liCA. And •.• LBDA is>^I3I>A, .-. Uii)A is greater than a nght angle, and the ^HCA is less than a right angle. "h"«- -.nlTr '^' i"?' ''""°" '° "><= «l"'->l"ies of the theorem, the angles C and C are both e<,ual to, or both greater or Loth less than a nght angle, the triangles are congruent. ^CA-Angles which are both greater than, or both equal to or both less than a right angle are said to be ./ M sZ .nl°; ^ w'r^^' '°"''''' "^ ''" P-'^'"*^' ^^'-^^ ^'^^^ '-^nd three angles. When two tr.angles are congruent all the parts in he one are respectively equal to the corresponding parts in he other. But in order to establish the congruence of two triangles ,t is not necessary to establish independently the respective equality of all the parts; for, as has now been shown ,f certain of the corresponding par's be equal the equality of the remaining parts and hence the congruence of the triangles follow as a consequence. Thus it is sufficient that two sides and the included angle in one triangle shall be respectively equal to two sides and the included angle in another. For, if we are given these parts, we are given con- sequentially all the parts of a triangle, since every trianHe havmg two sides and the included angle equal respectively^to those given is congruent with the given triangle. Hence a triangle is ,,^?7.r;z when two of it's sides and the angle between them are given. ii: F I 36 SYNTIIKTIC (;i<;OMETRV in mw A triangle is ^ive;i or determined by its elements being given according to the following table : — 1. Three sides, (eg") 2. Two sides and the included angle, (52°) 3. Two angles and the included side, (59°) 4. Two angles and an opposite side, (64°) 5. Two sides and the angle opposite the longer side, (65°) When the three parts given are two sides and the angle opposite the shorter side, two triangles satisfy the conditions, whereof one has the angle opposite the longer side supple- mentary to the corresponding angle in the other. This is known as the ambiguous case in the solution of triangles. A study of the preceding table shows that a triangle is completely given when any three of its six parts are given, with two exceptions : — (i) The three angles ; (2) Two sides and the angle opposite the shorter of the two sides. 67°. Theorem.— U two triangles have two sides in the one respectively equal to two sides in the other, but the included angles and the third sides unequal, then 1. The one having the greater included angle has the greater third side. 2. Conversely, the one having the greater third side has the greater included angle. A'B' = AB and B'C' = BC, and I. ^ABC is > z.A'B'C, then AC is > A'C. Then ABD is ABC in its new position. Proof.— Ltt A' be placed upon A and A'B' along AB. Since A'B' = AB, B' falls on B. Let C fall at some point D. ; solution of lorter of the TIIKKE OR MORE POINTS AND LINES. 17 (hyp.) (constr.) q.e.d. Let BE bisect the Z.DBC and meet AC in E. Join DE. Then, in the As DDE and CBE, DB=:BC, .lDBE = ^CBE, and BE IS common. ADBEsACBE, ''^"^ DE = CE. 15ut AC = AE + EC = AE + ED, which is greater than AD. ACis>A'C'. 2. AC is > A'C, then -ABC is greater than :_A'B'C'. Proof ~£\,^ proof of this follows from the Rule of Identity. (,^\ Thcorcm.~i. Every point upon a bisector of an an^rle IS equidistant from the arms of the angle. 2. Conversely, every point equidistant from the arms of an angle is on one of the bisectors of the angle. I. OP and 00 are bisectors of the angle AOB, and PA \ H are perpendiculars from P upon the arms. Then PA = PB. \ /^-TP Proof.-'X'a^ As POA and POB a^" ;u-e congruent, since they have two -^ /- angles and an opposite side equal in ^-' ^ each (64°); .-. PA = PB. If Q be a point on the bisector 00 it is shown in a similar manner that the perpendiculars from Q upon the arms of the angle AOB are equal. ^ ^ 2. If PA is _L to OA and PB is JL to OB, and PA=!pB then PO IS a bisector of the angle AOB. A-^./-The As POA and POB are congruent, since they hav^e two sides and an angle opposite the longer equal in each (65 , i) ; ... ^POA = ^POB, and PO bisects the lAOB. % iK' ■ 1.. m JM rlH m% lis 38 SYNTHETIC GEOMETRY. I! 'i Hi li Similarly, if the perpendiculars from Q upon OA ana OB are equal, (20 bisects the ^BOA', or is the external bisector of the ^OB. ^.^.^, LOCUS. 69°. A locus is the figure traced by a variable point, which takes all possible positions subject to some constraining condition. If the point is confined to the plane the locus is one or more lines, or some form of curve. Illustration.— \n the practical process of drawing a line or curve by a pencil, the point of the pencil becomes a variable (physical) point, and the line or curve traced is its locus. In geometric applications the point, known as \S\^ generat- ing point, moves according to some law. The expression of this law in the Symbols of Algebra is known as the equation to the locus. Cor. I. The locus of a point in the plane, equidistant from the end-points of a given line-segment, is the right bisector of that segment. This appears from 54°. Cor. 2. The locus of a point in the plane, equidistant from two given lines, is the two bisectors of the angle formed by the lines. This appears from 68°, converse. Exercises. f. How many lines at most are determined by 5 points.^ by 6 points .? by 12 points .? 2. How many points at most are determined by 6 lines ? by 12 lines .'' 3. How many points are determined by 6 lines, three of which pass through a common point } THREE OR MORE POINTS AND LINES. 39 us IS one or 4. How many angles altogether are about a triangle ? How many at most of these angles arc different in magni- tude ? What is the least number of angles of different magnitudes about a triangle ? 5. In Fig. of 53°, if Q be any point on PC, APAQ-APBQ. 6. In Fig. of 53°, if the Al'CB be revolved about PC as an axis, it will become coincident with APCA. 7. The medians to the sides of an isosceles triangle are equal to one another. 8. Prove 58° from the axiom "a straight line is the shortest distance between two given points." 9. Show from 60" that a triangle cannot have two of its angles right angles. 10. If a triangle has a right angle, the side opposite that angle is greater than either of the other sides. 11. What is the locus of a point equidistant from two sides of a triangle .'' 12. Find the locus of a point which is twice as for from one of two given lines as from the other. 13. Find the locus of a point equidistant from a given line and a given point. h"t SECTION IV. '** PARALLELS, ETC. 70°. Def.—Two lines, in the same plane, which do not intersect at any finite point 7\.xq paralld. Next to perpendicularity, parallelism is the most important directional relation. It is denoted by the symbol |1, which is to be read " parallel to " or " is parallel to " as occasion may reauire. 1 The idea of parallelism is identical with that of sameness „1| Ilii ij> m 40 SYNTHETIC GEOMETRY. iilr llliiblll I Ml ! II il 1 of direction. Two line-segments may differ in length or in direction or in both. If, irrespective of direction, they have the same length, they are equal ; if, irrespective of length, they have the same direction, they are parallel ; and if both length and direction are the same they are equal and parallel. Now when two segments are eg'uai one may be made to coincide with the other by superposition without change of length, whether change of direction is required or not. So when they are parallel one may be made to coincide with the other without change of direction, whether change of length is required or not. j4xwm.— Through, a given point only one line can be drawn parallel to a given line. This axiom may be derived directly from 24'. 71°. Theorevt. — Two lines which are perpendicular to the same line are parallel. L and M are both ± to N, then L is || to M. M Proof.— \i L and M meet at any point, two - perpendiculars are drawn from that point to the line N. But this is impossible (61°). Therefore L and M do not meet, or they are parallel. Cor. All lines perpendicular to the same line are parallel to one another. 72°. Thcorem.—Two lines which are parallel are perpen- dicular to the same line, or they have a common perpendicular. (Converse of 71".) L is II to M, and L is ± to N ; then M is ± to N. Proof. --U M is not ± to N, through any point P in M, let K be ± to N. line can be :ular to the PARALLELS, ETC. Then K is || to L. Kut M is II to L. Therefore K and M are both || to L, which is impossible unless K and M coincide. Therefore L and M are both ± to N, or N is a common perpendicular. 41 (71") (hyp.) (70", Ax.) "c/d e/f 7Z . Def.—h line which crosses two or more lines of any system of lines is a transversal. g Thus EF is a transversal to the lines . / ABandCD. ^^ — ^ In general, the angles formed by a transversal to any two lines are ~ ^/a distinguished as follows— /f a and e, c and^, b and/ d and h are pairs of corresponding angles. c and /; e and d are pairs of alternate angles. c and e, d and/are pairs of interadjacent angles. 74°. When a transversal crosses parallel lines— 1. The alternate angles are equal in pairs, 2. The corresponding angles are equal in pairs. 3- The sum of a pair of interadjacent angles is a straight angle. ^ AB is II to CD and EF is a transversal. A p / vk I. z.AEF = i.EFD. 3 ^^^^/— Through O, the middle point ^~ ^' of EF, draw PQ a common J. to AB and CD. Then AOPE^AOQF; ^eA " .. . ^AEF = ^EFD. ^.^.,/. Similarly the remaining alternate angles are equal. 2. iAEG = z.CFE, etc. Proof.— ^AEG = supplement of £AEF, (40°, Def. i) ^"^ ^CFE = supplement of lE FD. »1 \f,ii m •l!i ' 1 I' I ill" l.f 42 But SYNTHETIC GEOMETRV .aeg=.cfe; ^'\:;] Similarly the other corresponding angles are equal in pairs. 3. ^AEF+^CFE=_L. Proof.^ z.AEF = ^EFD, fy.o . and Z-CFE + z.EFD=_L; W) ^EF+^CFE=X. geJ. Cor. It is seen from the theorem that the equality of a pair of alternate angles determines the equality in pairs of corre- sponding angles, and also determines that the sum of a pair of interadjacent angles shall be a straight angle. So that the truth of any one of the statements i, 2, 3 determines the truth of the other two, and hence if any one of the statements be proved the others are indirectly proved also. 75°. Ty^^^A-^w.— If a transversal to two lines makes a pair of alternate angles equal, the two lines are parallel. (Con- verse of 74° in part.) If ^EF=^EFD, AB and CD are parallel. Proof.-DxTi.^ PQ as •„ ^^o^ j_ j^ ^p^^ AOPE^AOQF; (59°) •'• ^OPE=^OQF = "~I, and.-. AB is II to CD. {li'')q.e.d. Cor. It follows from 74° Cor. that if a pair of corresponding angles are equal to one another, or if the sum of a pair o> interadjacent angles is a straight angle, the two lines are parallel. 76°. Thcorem.-Th^ sum of the internal angles of a tri- ^B ^ angle is a straight angle. E ABC is a A; thez-A-j-^B+^C=J.. A C D D be any point on AC produced Proof. —"L^x. CE be !| to AB, and PARALLELS, ETC. 43 (74^ I) q.e.d. iqual in pairs. (74°, I) (38°) q.e.d. ility of a pair lirs of corre- jm of a pair So that the nes the truth atements be lakes a pair illel. (Con- \% X to AB, (59°) (71°)^.^.^. rresponding )f a pair of o lines are JS of a tri- o AB, and Then BC is a transversal to the parallels AB and CE ; andJt^X ^^^^^^ -"> or the .o.hns of a square are equa, and bisect eachtl^rr^hV^IL? transversal. insversal they do so upon every BF,-r ' ■.''""'"' AC = CE, and BF ^ any other transversal. Then BI) Then AGDC and CDHE are c=!s. Also, and and GD=AC=CE=DH ^GBD=^DFH,vAGis||toEF, z.BDG = z.FDH; ABDG^AFDH, BD = DF. A/-The figure ABFE is a /ra/>eso:W. inerefore a trapezoid is a nmHr-^^^i i, • ^ ^ ^^ ^ quadrangle having only (80°, 3) (81°, I) (74°, I) (40°) (64°) q.e.d. two t one another It angle is a equal is a he rhombus le another; s diagonals It angles. equal seg- pon every e parallels = CE, and Then BI) trough D (80°, 3) (81°, I) (74°, I) (40°) (64°) q.e.d. nly two PARALLELS, ETC. ^pr sides parallel. The parallel sides are the major and minor bases of the figure. Cor. I. Since (81°, 2) (74°, 3) ngles. qual to one 1 and 2CD=AG4-EH, =AB+BG+EF-HF BG=HF; CD = i(AB + EF). Or, the line-segment joining the middle points of the non- parallel sides of a trapezoid is equal to one-half the sum of the parallel sides. Cor. 2. When the transversals meet upon one of the extreme parallels, the figure AEF' becomes a A and CD' becomes a line passing through the middle points of the sides AE and AF', and parallel to the base EF'. Therefore, i, the line through the middle point of one side of a triangle, parallel to a second side, bisects the third side And, 2, the line through the middle points of two sides of a triangle is parallel to the third side. 85°. 77/^^rm.-The three medians of a triangle pass through a common point. CF and AD are medians intersecting in O. Then BO is the median to AC. Proof.—l^^t BO cut AC in E, and let AG II to FC meet BO in G. Join CG. Then, BAG is a A and FO passes through the middle of AB and is || to AG, ••. O is the middle of BG. (84'', Cor 2) Again, DO passes through the middle points of two sides of the ACBG, CG is II to AO or OD ; AOCG is a ^"d AE = EC; BO is the median to AC. (84°, Cor, 2) (81°, 3) q.e.d. \i\ ,f j I .| i\ !"'■' 48 SYNTHETIC GEOMETRY. i.j|j .n,i .i ^^/-When three or more lines meet in a point they are said to be concurrent. Therefore the three medians of a triangle are concurrent. Def 2.-The point of concurrence, O, of the medians of a triangle is the centroid of the triangle. ofnTof^OG^ ^ '' '^^ '^'^^^'' ^""'"^ °^^^'' '''"'^ ^ '' ^^^ ""'^^^^ OE = ^OB, ^^^°'^^ Therefore the centroid of a triangle divides each median at two-thirds of its length from its vertex. 86°. TheorcuL~T\,^ three right bisectors of the sides of a triangle are concurrent. Proof.-'LQi L and N be the right bisectors of BC and AB respectively. Then L and N meet in some point O. c- T • , . , , . (79°> Cor.) Since L is the right bisector of BC, and N of AB O is equidistant from B and C, and is also equidistant from A ^"^ ^- (53°) Therefore O is equidistant from A and C, and is on the right bisector of AC. / .. Therefore the three right bisectors meet at O. g^e.d. Cor. Since two lines L and N can meet in only one point (24°, Cor. 3), O is the only point in the plane equidistant from A, B, and C. Therefore only one finite point exists in the plane equi- distant from three given points in the plane. Z?^/-The point O, for reasons given hereafter, is called the circumccntre of the triangle ABC. Zf, Z)^/— The line through a vertex of a triangle per- pendicular to the opposite side is the perpendicular to that side, and the part of that line intercepted within the triangle is the altitude to that side. PARALLELS, ETC. 49 Where no reference to length is made the word altitude is often employed to denote the indefinite line forming the perpendicular, Hence a triangle has three altitudes, one to each side. 88°. Theorem.— i:\i^ three altitudes of a triangle are con- current. /V^^/_Let ABC be a triangle. Complete the mo^, ACBF, ABDC, and ABCE. Then •.• P'^B is || to AC, and BD is || to AC, FBD is one line, (70°, Ax.) ''^nd FB = BD. (81°, I) Similarly, DCE is one line and DC = CE, and EAF is one line and EA = AF. Now, •.• AC is II to FD, the altitude to AC is J. to FD and passes through B the middle point of FD. (72°) Therefore the altitude to AC is the right bisector of FD, and similarly the altitudes to AB and BC are the right bisec- tors of DE and EF respectively. But the right bisectors of the sides of the ADEF are concurrent (86°), therefore the altitudes of the AABC are concurrent. ^,^_ ^^Z— The point of concurrence of the altitudes of a tri- angle is the orthocentre of the triangle. Cor. I. If a triangle is acute-angled {-jf, 5), the circum- centre and orthocentre both lie within the triangle. 2. If a triangle is obtuse-angled, the circumcentre and orthocentre both lie without the triangle. 3- If a triangle is right-angled, the circumcentre is at the middle point of the side opposite the right angle, and the orthocentre is the right-angled vertex. -^^— "The side of a right-angled triangle opposite the ri"-ht angle is called the hypothcnuse. D fffi 50 SVNTIIKTTC GEOMETRY. 89°. The definition of 80° admits of three different figures VIZ, : — '^ ' I. The normal quadrangle (i) in which each of the in- ternal angles is less than a straight angle. When not (0 otherwise qualified the term quadrangle will mean this figure. 2. The quadrangle (2) in which one of the internal angles as at D, is greater than a straight angle. Such an angle in a closed figure is called a re-entrant angle. We will call this an inverted quadrangle. 3. The quadrangle (3) in which two of the sides cross one another. This will be called a crossed quadrangle. In each figure AC and BD are the diagonals T so that both diagonals are within in the normal quadrangle, one is within and one without in the inverted quadrangle, and both are without in the crossed quadrangle. The general properties of the quadrangle are common to all three forms, these forms being only variations of a more general figure to be described hereafter. 90°. Theorem.— ThQ sum of the internal angles of a quad- rangle is four right angles, or a circumangle. Proof.— 'Y\vQ angles of the tM'o As ADD and CBD make up the mternal angles of the quadrangle. But these are J_ + J_ ; (76°) therefore the internal angles of the quadrangle are together equal to four right angles. ^^^^ Cor. This theorem applies to the inverted quadrangle as is readily seen. m 1 »ri PARALLELS, ETC. 51 91 . 7y/.m.;«.-If two lines be respectively perpendicular to two other lines, the angle between the first two is equal or supplementary to the angle between the last two. BC is J_ to AR and CD is J_ to AD. Then l^BC . CD) is equal or supplemen- tary to ^(AB. AD). Proo/.~ABCD is a quadrangle, and the ^s at B and D are right angles : z.BAD+z.BCD = x, °^ ^BCD is supplementary to ^BAD. But ^BCD is supplementary to ^ECD ; and the ^(BC. CD) is either the angle BCD or DCE. l{BC . CD) is = or supplementary to ^BAD. (Iiyp.) (90°) (39°) Exercises. 1. ABC IS a A, and A', B', C are the vertices of equilateral As described outwards upon the sides BC, CA, and AB respectively. Then AA'=BB' = CC'. (Use 52°.) 2. Is Ex. I true when the equilateral As are described " mwardly " or upon the other sides of their bases ? 3. Two lines which are parallel to the same line are parallel to one another. h L' and M' are two lines respectively parallel to L and M TheL{L\M')=L(L.M). ;. On a given line only two points can be equidistant from a given point. How are they situated with respect to the perpendicular from the given point ? ). Any side of a A is greater than the difference between the other two sides. The sum of the segments from any point within a A to the three vertices is less than the perimeter of the A i>. ABC is a A and P is a point within on the bisector of ^. Then the difference between PB and I C is less than that between AB and AC, unless the A is isosceles 7. '^ -l > ;> k^^^-^ If I f 1 1 'f I kt i i 1 W: !) J 52 9- lo. 1 1. 12. U- 14. '5- 1 6. 17. 18. 19. 20. 21. SYNTHETIC GEOMETRY. Is Ex. 8 true when the point V is without the A. ^ut on the same bisector .-* Examine Ex. 8 when P is on the external bisector of A, and modify the wording of the exercise accordingly. CE and CF are bisectors of the angle between AH and CU, and EF is parallel to AH. Show that EF is bisected by CD. If the middle points of the sides of a A be joined two and two, the A 's divided into four congruent As. From any point in a side of an equilateral A lines arc drawn parallel to the other sides. The perimeter of the £1117 so formed is ecjual to twice a side of the A- Examine Ex. 13 when the point is on a side pro- duced. The internal bisector of one angle of a A and the ex- ternal bisector of another angle meet at an angle which is equal to one-half the third angle of the A- O is the orthocentre of the AABC Express the angles AOB, BOC, and COA in terms of the angles A, B, and C. P is the circumcentre of the AABC. Express the angles APB, BPC, and CPA in terms of the angles A, B, and C. The joins of the middle points of the opposite sides of any quadrangle bisect one another. The median to the hypothenuse of a right-angled triangle is equal to one-half the hypothenuse. If one diagonal of a OZJ be equal to a side of the figure, the other diagonal is greater than any side. If any point other than the point of intersection of the diagonals be taken in a quadrangle, the sum of the line-segments joining it with the vertices is greater than the sum of the diagonals. If two right-angled As have the hypothenuse and an acute angle in the one respectively equal to the like parts in the other, the As are congruent. PAKALIJ.LS, ETC 53 ^3- 24. 27. 28. 29. The bisectors of two adjacent angles of a / — 7 are _L to one another. AliC is a A. The angle between the external bisector of B and the side AC is A(C~A). The external bisectors of B and C meet in D. Then z.BDC=--i(B + C). A line L v/hich coincides with Cm side AB of the AABC rotates about B until it coincides with BC, without at any time crossing the triuigle. Through what angle does it rotate ? The angle required in Ex. 26 is an external angle of the triangle. Show in this way that the sum of the three external angles of a triangle is a circumanglc, and that the sum of the three internal angles is a straight angle. What property of space is assumed in the proof of Ex. 27? Prove 76° by assuming that AC rotates to AB by crossing the triangle in its rotation, and that AB rotates to CB, and finally CB rotates to CA in like manner. i^/ 2.— The segment of a secant included within the is a chord. Thus the line L, or AB, is a secant, and the segment AB is a chord. (21°) The term chord whenever involv- -s^.__^^ ing the idea of length means the segment having its end- points on the circle. But sometimes, when length is not involved, It IS used to denote the whole secant of which it properly forms a part. ^^/ 3--A secant which passes through the centre is a centre-line, and its chord is a diameter. Where length is not implied, the term diameter is some- times used to denote the centre-line of which it properly forms a part. ^ ^ Thus M is a centre-line and CD is a diameter. 96°. 77/6'^m/A— Through any three points not in line- 1. One circle can be made to pass. %s> 2. Only one circle can be made to pass. Proof.-h^t A, B, C be three points not in line. Join AB and BC,and let L and M be the a right bisectors of AB and BC respectively. I. Then, because AB and BC intersect at iJ, L and M intersect at some point O, (79°, Cor.) f.l Ill 56 SVN'lJILTJC GliOMETKV. 1 . i,i and O is equidistant from A, B, and C. (86'^) .•• the with centre at O, and radius equal to OA, passes through B and C. y g.e.d. 2. Any through A, B, and C must have its centre equally distant from these three points. But O is the only point in the plane equidistant from A, B and C. -„,„ ' ' A A , (86°, Cor.) And we cannot have two separate 0s having the same centre and the same radius. /^.o . .'. only one circle can pass through A, B, and C. q'.c.d. Cor. r. Circles which coincide in three points coincide altogether and form one circle. Cor. 2. A point from which more than two equal segments can be drawn to a circle is the centre of that circle. Cor. 3. Since L is a centre-line and is also *he ri^ht bisector of AB, ' ^ .-. the right bisector of a chord is a centre iine. Cor. 4. The AAOB is isosceles, since OA=OB Then if D be the middle of AB, OD is a median to the base AB and is the right bisector of AB. /.r- q^^ ^-v .-. a centre-line which bisects a chord is perpendicular t"o the chord. Cor. 5. From Cor. 4 by the Rule of Identity, A centre line which is perpendicular to a chord bisects the chord. .-.the right bisector of a chord, the centre-line bisectino- the chord, and the centre-line perpendicular to the chord are one and the same. 97°. From 92°, Def , a circle is given when the position of its^centre and the length of its radius are given. And, from 96°, a circle is given when any three points on it are given. It will be seen hereafter that a circle is determined by three points even when two of them become coincident, and in higher geometry it is shown that three points determine a THE CIRCLE. rj circle, under certain rircumstances, when all three of the points become coincident. Def.—Any number of points so situated that a circle can pass through them are said to be coHcyclk, and a rectilinear figure (14°, Def.) having its vertices concyclic is said to be mscnbed in the circle which passes through its vertices, and the circle is said to circumscribe the figure. Hence the circle which passes through three given points is the circuvicirclc of the triangle having these points as ver- tices, and the centre of that circle is the circiuncentre of the triangle, and its radius is the circnmradius of the triangle. A like nomenclature applies to any rectilinear figure having its vertices concyclic. 98°. T/tcorcm.—U two chords bisect one another they are both diameters. If AP=:PD and CP = PB, then P is the '^^ ^3 centre. Proo/.~Since P is the middle point of both AD and CB (hyp.), therefore the right bisectors of AD and CB both pass through P. But these right bisectors also pass through the centre; (96°, Cor. 3) .-. p is the centre. (24°, Cor. 3) ^.e.J. 99°- T/icorcm.— Equal chords are equally distant from the centre ; and, conversely, chords equally dis- tant from the centre are equal. . -t~~-o If AB = CD and OE and OF are the per- pendiculars from the centre upon these chords, then OE = OF; and conversely, if OE = OF, then AB = CD. c- Proo/.—Since OE and OF are centre lines 1 to AB and CD, AB and CD are bisected in E and F. (96°, Cor. 5) ••- in the As OBE and ODF OB-OD, EB = FD, .■( Wi , IJ 58 SYNTHETIC GEOMETRY. loo A and they are right-angled opposite equal sides, AOBE=AODF, and OE = OF. Conversely, by the Rule of Identity, if OE = OF then AB = CD. ' , q.e.a. Theoran.— Two secants which make equal chords B -pmake equal angles with the centre-line through their point of intersection. AB = CD, and PO is a centre-line through the point of intersection of AB and CD. Then ^PO=z.CPO. Profl/.~het OE and OF be ± to AB and CD from the centre O. Then OE = OP^, (99") AOPE^AOPF, (65°) and ^APO=^CPO. g.c.d. Cor. I. •.• E and F are the middle points of AB and CD, (96°, Cor. 5) ••• PE = PF, PA = PC, and PB = PD. Hence, secants whici. make equal chords make two pairs of equal line-segments between their point of intersection and the circle. Cor. 2. From any point two equal line-segments can be drawn to a circle, and these make equal angles with the centre-line through the point. 101°. As all circles have the same form, two circles which have equal radii are equal and congruent (93°, 4), (51°). Hence equal and congruent are equivalent terms' when applied to the circle. Lfef. r.— Any part of a circle is an arc. The word equal when applied to arcs means congruence or capability of superposition. Equal arcs come from the same circle or from equal circles. 1 ! i . f THE CIRCLE. 59 Def. 2.— A line which divides a figure into two parts such that when one part is revolved about the line it may be made to fall on and coincide with the other part is an axis of symvietry of the figure. 1 02 Theorem.— K centre-line is an axis of symmetry of the circle. Proof.— Lei AB and CD be equal chords meeting at P, and let PHOG be a centre line. Let the part of the figure which lies q\ upon the F side of PG be revolved about PG until it comes to the plane on the E side of PG. Then-,- ^GPA = z.GPC, (roo") .-. PC coincides with PA. And -.- PB = PD and PA=PC, (100", Cor. i) .'. U coincides with B, and C coincides with A. And the arc HCG, coinciding in three points with the arc HAG, is equal to it, and the two arcs become virtually but one arc. (96°^ Cor. i) Therefore PG is an axis of symmetry of the 0, and divides it into two equal arcs. q.e.d. Z) If 62 SYNTHETIC GEOMETRY. i-^ III t breaks or interruptions ; or, a generating point in passing from one position to another must pass through every inter- mediate position. 2. In Art. 53° we have the theorem— Every point on the right bisector of a segment is equidistant from the end-points of the segment. In this theorem the limiting condition in the hypothesis is that the point must be on the right bisector of the segment. Now, if P be any point on the right bisector, and we move P along the right bisector, the limiting condition is not at any time violated during this motion, so that P remains con- tinuously equidistant from the end-points of the segment during its motion. We say then that the property expressed in the theorem is continuous while P moves along the right bisector. 3. In Art. 9^° we have the theorem— The sum of the in- ternal angles of a quadrangle is four right angles. The limiting condition is that the figure shall be a quad- p^ rangle, and that it shall have in- ternal angles. Now, let ABCD be a quadrangle. Then the condition is not violated if D moves to D^ or Dg. But in the latter case the normal quad- rangle ABCD becomes the inverted quadrangle ABCD2, and the theorem remains true. Or, the theorem is continu- ously true while the vertex D moves anywhere in the plane, so long as the figure remains a quadrangle and retains four internal angles. Future considerations in which a wider meaning is given to the word " angle " will show that the theorem is still true even when D, in its motion, crosses one of the sides AB or BC, and thus produces the crossed quadrangle. The Principle of Continuity avoids the necessity of proving theorems for different cases brouo-ht about b^'' variations in the disposition of the parts of a diagram, and it thus gener- THE CIRCLE. 63 ahzes theorems or relieves them from dependence upon the particularities of a diagram. Thus the two figures of. Art 100 differ in that in the first figure the secants intersect without the circle, and in the second figure they intersect within, while the theorem applies with equal generality to both. The Principle of Continuity may be stated as follows •— When a figure, which involves or illustrates some geometric property, can undergo change, however small, in any of its parts or in their relations without violating the conditions upon which the property depends, then the property is con- tinuoHs while the figure undergoes any amount of change of the same kind within the range of possibility. 105°. Let AB be a chord dividing the into unequal arcs, and let P and Q be any points upon the major and minor arcs respectively. T .r^u u (102°, Def.) Let O be the centre. I. The radii OA and OB fonn two angles at the centre, a major angle denoted by a and a minor angle de- noted by ^A These together make up a circumangle. 2. The chords PA, PB, and QA, QB form two angles at the circle, of which APB is the ininor angle and AQB is the major angle. 3. The minor angle at the circle, APB, and the minor angle at the centre, /3, stand upon the minor arc, AQB, as a base Similarly the major angles stand upon the major arc as base' 4. Moreover the ^APB is said to be /;/ the arc APB so that the minor angle at the circle is in the major arc, and 'the major angle at the circle is in the minor arc. 5- When B moves towards B' all the minor elements increase and all the major elements decrease, and when B comes to B' the minor elements become respectively equal to the major, and there is neither major nor minor ('Hi Ill 4.!. 51' [' 64 SYNTHETIC GEOMETRY. When B, moving in the same direction, passes B', the elements change name, those which were formerly the minor becoming the major and vice versa. Io6^ Theorem. — An angle at the circle is one-half the corresponding angle ?.t the centre, major corresponding to major and minor to minor. p Z.AOB minor is 2Z.APB. Proof. — Since A^PO is isosceles, i.OAP = z.OPA,(53°,Cor.i) and_OAP + ^OPA = 2^0PA. But ^AOC=lOAP + ^OPA, (76^ Cor.) z.AOC = 2Z.OPA. Similarly ^B0C = 2^0FB ; .".adding, Z.AOB minor = 2^PB. q.e.d. The theorem is thus proved for the minor angles. But since the limiting conditions require only an angle at the circle and an angle at the centre, the theorem remains true while B moves along the circle. And when B passes B' the angle APB becomes the major angle at the circle, and the angle AOB minor becomes the major angle at the centre. the theorem is true for the major angles. Cor. I. The angle in a given arc is constant. (105°, 4) Cor. 2. Since _APB = |^z.AOB minor, and ^AQB = I^AOB major, and •.' lAOB minor + Z.AOB major = 4 right angles (37°) Z.APB +_vQB = a straight angle. And APBQ is a concyclic quadrangle. Hence a concyclic qu.idrangle has its opposite internal angles supplementary. (40°, Det. i) Cor. 3. D being oii AQ produced, iiHQD is supplementary to Z-AOB. mh THE CIRCLK. 65 Hut Z.APB is supplementary to Z.AQU, ^APB = ^1}(MJ. Hence, if one side of a concyclic quadrangle be produced, the external angle is equal to a the opposite internal angle. Cor, 4. Let B come to B'. (Fig. of 106°) Then ^AO B' is a straight angle, ^APB' IS a right angle. But the arc APB' is a semicircle. Q\ \0 (102°, Def ) Therefore the angle in a semicircle is a right angle 107°. Theorem.~A quadrangle which has its opposite angles supplementary has its vertices concyclic. (Converse of 106°, Cor. 2) ABCD is a quadrangle whereof the zADC is supplementary to ^BC ; then a circle can pass through A, B, C, and I). Proof.— U possible let the through A, B, and C cut AD in some point P. Join P and C. Then ^APC is supplementary to Z.ABC, (106°, Cor. 2) and lADC is supplementary to Z.ABC, z.APC-^DC, which is not true. .-. the cannot cut AD in any point other than D, Hence A, B, C, and D are concyclic. Cor. I. The hypothenuse of a right-angled triangle is the diameter of its circumcircle. (88°, 3, Def. ; ()f, Def.) Cor. 2. When V moves along the the AAPC (last figure) has its base AC onstant and its vertical angle APC constant. Therefore the locus of the vertex of a triangle which has a constant base and a constant vertical angle is an arc of a cir le passing through the end-points of the base. This property is employed in the transmel which is used to describe an arc of a given circle. £ (hyp.) (60°) g.c.d. m \\ if. i t %■ . " '''1 !] 66 S Y N Til KTIC G K(3M ETR Y. ill i b It consists of two rules (i6 ) L and M joined at a L ^-^ determined an},de. When it is made to slide over two pins A and H, a pencil at I' traces an arc passing through A and B. io8\ 7Vicofi'//i.—The angle between two intersecting se- cants is the sum of those angles in the circle which stand on the arcs intercepted between the secants, when the secants intersect within the circle, and is the difference of these angles when the secants intersect without the circle. _APC-_ABC + _BCD, (,!,^0 _APC = _ABC-_BCD. (}.!!;!.) Proof.-\. -APC = z_PBC-f ^PCB, (60°) .-. ^APC = _ABC + _BCD. 2. _ABC = _APC + _BCP, .-. ^APC = _ABC-_BCD. q.e.d. EXKRCISES. 1. If a six-sided rectilinear figure has its vertices concyclic, the three alternate internal angles are together equal to a circumangle. 2. In Fig. 105°, when B comes to O, BQ vanishes ; what is the direction of BO just as it vanishes ? 3. Two chords at right angles determine four arcs of which a pair of opposite ones are together equal to a semi- circle. 4. A, B, C, D are the vertices of a square, and A, E, F of an equilateral triangle inscribed in the same circle. What is the angle between the lines BE and DF.? between BF and ED ^ TllK CIRCLi:. 67 SPFXIAL SECANTS-TANGENT. ^^^^^09^ Let P be a Hxed point on the 0S and O a variable The position of the secant L, cut- ting the circle in P and Q, depends upon the position of O. As Q moves alon- the the secant rotates about P as pole. While Q makes one complete revolution along the the secant L passes through two ^ special positions. The fir^f of fi-.«.^ • 1 I- r ^^ *^' tnesc IS when () ^s fnithr.cf ; .stant fron, P, as at Q , and the secant L becomes at '"' ZJ " "*'" " "'"^^ "'"> ™inci/;// o/.oufaa Bemg formed by the union of two points it iCre- sents both, and is therefore a c/oM point. ^ From Defs. i and 2 we conclude— T. A point of contact is a double point. 2. As a line can cut a only twice it can touch a only once 3. A hne which touches a cannot cut it. 4. A is determined by two points if one of them is a given pomt of contact on a given line ; or, only one arcle can pass through a given point and touch a given hne at a given point. (Compare 97° ) i t 6^ SYNTHETIC GEOMETRY iio°. Thcoi'em. — A centre-line and a tangent to the same point on a circle are perpendicular to one another. L' is a centre-line and T a tan- gent, both to the point P. Then L' is i to T. Proof. — *.• T has only the one point P in common with the 0, every point of T except P lies without the 0. .' . "Y' if O is the centre on the line L', OP is the shortest segment from O to T. OP, or L', is ± to T. (63°, i) qu\d. Cor. I. Tangents at the end-points of a diameter are parallel. Cor. 2. The perpendicular to a tangent at the point of con- tact is a centre-line. (Converse of the theorem.) Cor. 3. The perpendicular to a diameter at its end-point is a tangent. 111°. Theorem. — The angles between a tangent and a chord from the point of contact are respectively equal to the angles in the opposite arcs into which the chord divides the circle. TP is a tangent and PO a chord to the same point P, and A is any point on the 0. Then ^QFT=lOAP. Let PD be a diameter. .lQAP=^QDP, ^DQP is a ~1- ^DPO is comp. of ^QPT, Z.DPQ is comp. of -QDP, ^0DP=40PT==QAP, Similarly, the .iQPT'=, .'. two circles can intersect in only two points. Cor. Two circles can touch in only one point. For a pomt of contact is equivalent to two points of intersection. 1 13°. Theorem. ~T\\Q common centre-line of two intersect- mg circles is the right bisector of their common chord. O and O' are the centres of S and S'j and AB is their common chord, bisector of AB. Then 00' is the right Proof.- and Similarly Since AO = BO, AO' = BO', • •. O is on the right bisector of AB. O' is on the right bisector of A^, .". 00' is the right bisector of AB. (54°) Cor. r. By the principle of continuity, OO' always bisects A 15. Let the circles separate until A ar d I] coincide. Then the circles touch and OO' passes through the point of contact. Def.—TwQ circles which touch one another have external contact when -ach- circle lies without the other, and internal contact when one circle lies within the other. Cor. 2. Since OO' (Cor. i) passes through the point of contact when the circles touch one another — {a) When the distance between the centres of two c.rcles is the sum of their radii, the circles have external contact. Kb) When the distance between the centres is the difference of the radii, the circles have internal contact. 'M fflil t 70 SYNTHETIC (rKOMKTKV (0 When the distance between the centres is «;reater than the sum of the radii, the circles exchide each other without contact. {d) When the distance between tlic centres is less than the difft:rence of the radii, the greater circle includes the smaller without contact, (f) When the distance between the centres is less than the sum of the radii and greater than their difference, the circles intersect. ilii 114°. 77ic-c»Ym. —Vxom any point without a circle two tangents can be drawn to the circle. /'roof.— Let S be the and V the point. Upon the segment PO as diameter let the 0S' be described, cutting 0S in A and 15. Then I'A and PH are both ' tangents to S. For Z.OAP is in a semicircle and is a "~|, (106°, Cor. 4) •■• ^P is tangent to S. (j jq", Cor. 3) Similarly DP is tangent to S. Cor. I. Since PO is the right bisector of AB, (113') PA=PB. (^530^ Hence calling the segment l^A the tanocnt from P to the circle, when length is under consideration, we have— The two tangents from any point to a circle are equal to one another. Def.~i:\\Q line AB, which passes through the points of contact of tangents from P, is called the chord of contact for the point P. w-f. Def I —The angle at which two circles intersect is the angle between their tangents at the point of intersection. Def 2. When two circles intersect at riglu. angles they are said to cut each other orthogonally. Tin: CIRCLI-: ;i The same term is conveniently applied to the intersection of any two figures at riglit angles. Cor. I. If, in the Fig. to 1 14^ PA be made the radius of a ( n-cle and P its centre, the circle will cut the circle S ortho- gonally. For the tangents at A are respectively perpendicular to the radii. Hence a circle S is cut orthogonally by any circle having Its centre at a point without S and its radius the langen't from the point to the circle S. Fr6°. The following examples furnish ti-orems of some importance. Ex. I. Three tangents touch the circle S at the points A, B, and C, and inter- sect to form the AATi'C. O ])eing the centre of the circle, ^A0C = 2_A'0C'. Proof.— AC' = BC', and BA' = CA', (114°, Cor. i) AAOC'^ABOC, and ABOA' = ACa^ z.BOC' = _AOC', and ^COA'=^BOA' Z-A0C = 2_A'0C'. ' ^^ Similarly z.A0B-2_A'OB', and ^BOC-2_B'OC'. ^^ If the tangents at A and C are fixed, and the tangent at B IS variable, we have the following theorem :-- The segment of a variable tangent intercepted bv two fixed tangents, all to the same circle, subtends a fixed angle at the centre. Kx. 2. If four circles touch two and two externallv, the points of contact are concyclic. I.et A, B, C, D be the centres of the circles, and P, f ), R, S be the points of contact. ' "' Then AB passes through P, BC through O, etc. (i 13", Cor. i) . I IM yy li " 1 Vi 'L n SYNTH lOriC CJ'OMKTRY. I i | their difference, the circles will meet in two points P and O. (113°, Cor. 2, e) The line PQ is the right bisector required. Proof.—? and O are each equidistant from A and B and .'. they are on the right bisector of AB ; (54") .*. PQ is the right bisector of AB. Cor. I. The same construction determines C, the middle point of AB. Cor. 2. If C be a given point on a line, and we take A and B on the line so that CA = CB, then the right bisector of the segment AB passes through C and is ± to the given line. .*. the construction gives the perpendicular to a given line at a given point in the line. 120°. Problem. — To draw a perpendicular to a given line from a point not on the line. al. rf. ^L 78 SYNTHETIC GEDMKTRY. Let be the -iven line and V be the point. Co/is/r.~ Draw any line through 1' meeting L at some point A. Hisect AP in C (119°, Cor. i), and with C as centre and CP as radius describe a circle. If PA is not J. to L, the will cut L in two points A and D. Then PD is the _L required. Proof.- VD A is the angle in a semicircle, -PDA is a-]. (106; Cor. 4) Cor. Let I) be a given point in L. With any centre C .md CD as radius describe a circle cutting L again in some ponit A. Draw the radius ACP, and join D and P 'j-Jicn DPis_LtoL. .-. the construction draws a ± to L at a given point in L. (Compare ik/, Cor. 2) Cor. 2. Let L be a given line and C a given point. To draw through C a line parallel to L. With C as the centre of a circle, construct a figure as given. Bisect PD in E (119^ Cor. i). Then CE is^|| to L tor C and E are the middle points of two sides of a trian-dc of which L is the base. /q,o n \ (64 , Cor. 2) 771, S(;uan^.— The square consists of two rules with their edges fixed permanently at right angles, or of a triangular plate of wood or metal having "^ y two of its edges at right angles. "^ ^— ^ To test a square. Draw a line AB and place the square as at S, so that one edge coincides with the line, and along the other edge draw the line ^ CD. Next place the square in the position S'. If the edges can 121 CONSTRUCTIVE GEOMETRY. 79 now be made to coincide ^v\^h the two lines the square is true. This test depends upon the f.ct that a right angle is one- half a straight angle. The square is employed practically for drawing a line ± to another line. Cor. I. The square is employed to draw a series of parallel lines, as in the figure. Cor. 2. To draw the bisectors of an angle by means of the scjuare. Let AOB be the given angle. Take OA=OB, and at A and B draw perpendiculars to OA and OB. Since AOB is not a straight angle, these perpendiculars 1 lien OC is the mternal bisector of _A0]1 For the tri- angles AOC and BOC are evidently congruent. -aoc=_b6c. The line drawn through O ± to OC is the external bisector. 122°. /V^/V.w.-Through a given point in a line to draw a line which shall make a given angle b with that line. Let P be the given point in the line L, and let X be the given '^^ angle. Constr— From any point B in ^ a' l the arm OB draw a ± to the arm OA. (i''o°) Afake PA'-OA, and at A' draw the ± A'B' makin- A'B' = AB. PB' is the line required. /'myi—The triangles OBA and PB'A' are evidently con- gruent, and .-. ^BOA=:X = _B'PA'. Cor. Since PA' might have been taken to the left of P, the problem admits of two solutions. When the angle X 'is a right angle the two solutions become one. i 4< % ill - e>. A ,0. IMAGE EVALUATION TEST TARGET (MT-S) 1.0 ISO 128 1^ 1^ M 2.2 I.I us u 11.25 2.0 18 i^ Hi 1.6 Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. KS80 (716) 872-4503 iV i V ^9) V m - ^ ^\ "O^ 6^ 8o SYNTHETIC GEOMETRY. The Protractor. — This instrument has different forms depending upon the accuracy re- quired of it. It usually consists of a semicircle of metal or ivory divided into degrees, etc. (41°). The point C is the centre. By placing the straight edge of the instrument in coincidence with a given line AB so that the centre falls at a given point C, we can set off any angle given in degrees, etc., along the arc as at D. Then the line CI) passes through C and makes a given angle with AB. 124 Prooletn. — Given the sides of a triangle to construct it. Constr. — Place the three sides of the triangle in line, as AB, BC, CD. With centre C and radius CD describe a circle, and with centre B and radius BA describe a circle. Let E be one point of intersection of these circles. Then AI^EC is the triangle required. Proof.— WE^B^ and CE = CD. Since the circles intersect in another point E', a second triangle is formed. But the two triangles being congruent are virtually the same triangle. Cor. I. When AB = BC = CA the triangle is equilateral. (53°, Def. 2) In this case the circle AE passes through C and the circle DE through B, so that B and C become the centres and BC a common radius. Cor. 2. When BC is equal to the sum or difference of AB and CD the circles touch (113°, Def) and the triangle takes the limiting form and becomes a line. When BC is greater than the sum cr less than the differ- ence no tria The each c differej 125°. by sup] backwa amongs the con This of equa be mad( Ther 126°. and the Let a the medi Suppo triangle median. By CO! and join equal to ( the trian: and is coi Thence Cor. Si possibility therefore sum, and niinous sic CONSTRUCTIVE GEOMETRY- gl r^i'^eiriT^r ^'-'-<'° - -' ('-3", Def.) and difference of the othe? two '"' ^^'^'" *^" ">« : t I backwards from the comDletrH fi ''^ reasoning amongst the given LrKh ^f '° '""^ ^^'^«°n the construction "^ "^ "'"'' °^ "■'"^'' "« "^^^ "^ke bemlderuTtttt'sC """ '"^"^' '^'"^"^^""' The next three problems furnish examples anfIe:^ittrtI:t=e'are'S^ "^ '^ ^^ Let « and ^ be two sides and « the median to the third side. Suppose ACB is the required triangle having CD as the given median. By completing the CZ7ACBC' and joining DC, we have DC c equal to CD and in the same line, and BC=Ar r«r°^ a the triangle CCB has CC-... nx, ^^ ^' ^"^ and is constructed by 124°. ' ^ =AC=^, Thence the triangle ACB is readily constructed. J iiic iriangie ALB depends upon that of CC'n ■ refore a median of a triangle is less 'than onel^f'^th; mi^s sii::"" *■''" ""^■'"^'^ *« '■"r-"- "f 'he contt' p (124°, Cor. 2) 82 SYNTHETIC GEOMETRY. I m il :; 127°. Problem.—To trisect a given line-segment, />., to divide it into three equal parts. Cofis/K— Let AB be the segment. '--..^^ Through A draw any line CD and make " 'B AC=AD. Bisect DB in E, and join CE, ^E cutting AB in F. Then AF is ^AB. Proo/.~CBD is a A and CE and BA are two medians. AF=^AB. (85°, Cor.) Bisecting FB gives the other point of division. -To construct a A when the three medians are given. Let /, m, 71 be the given medians, and suppose ABC to be the required triangle. Then AD = /, BE = ?«, and CF=«, AO = f/, OB = §w, and OF = i;/, .•. in the AAOB we have two sides and the median to the third side given. Thence AAOB is constructed by 126° and 127°. Then producing FO until 0C = 2F0, C is the third vertex of the triangle required. Ex. To describe a square whose sides shall pass through four given points. Let P, Q, R, S be the given points, and suppose ABCD to be the square required. Join P and Q upon opposite sides of the square, and draw QG || to BC. Draw SX ± to PQ to meet BC in E, and draw EF II to CD. ThenAQPG^AFSE, and SE = PQ. Hence the construction : — Join any two points PC), and through a third point S draw CONSTRUCTIVE GEOMETRY. 83 SX ± to PQ. On SX take SE = PQ and join E with the fourth point R. ER is a side of the squari in posidon .nd direction, and the points first joined, P 'and Q, are Topno site sides of the required square. ^^ Thence the square is readily constructed Since SE may be measured in two directions along the me SX, two squares can have their sides passing through the same four points P, Q, R, s, and having P and Q on opposite sides. ^ -^ ^^ Also, since P may be first connected with R or S, two hlvinTp'^/n 'T^''' '"^'"^"^^ ^^^ conditions aid having P and Q on adjacent sides. Therefore, four squares can be constructed to have their sides passing through the same four given points CIRCLES FULFILLING GIVEN CONDITIONS. The problems occurring here are necessarily of an elemen- at LlLn'S:-^" ''-'-''' ' ^'^* '° 'o-'' ^ Si-" '- P is a given point in the line L. G?;w/r.— Through P draw M _L to L. A circle having any point C, on M, as centre and CP as radius touches L at P. Proo/.~L is ± to the diameter at its " end-point, therefore L is tangent to the circle, (i 10°, Cor. 3) D,/.-As C is anj^ point on M, any number of circles may be ^awn to touch L at the point P, and all their centres lie Such a problem is Me/iu^V. because the conditions are not sufficient to determine a Jxrn/cu/ur circle. If the circle 84 SYNTHETIC GEOMETRY. ttm. varies its radius while fuWIling the conditions of the problem the centre moves alon^r m ; and M is called the avz/n'- W of the variable circle. Hence the centre-locus of a circle which touches a fixed line at a fixed point is the perpendicular to the line at that point. Cor. If the circle is to pass through a second given point Q the problem is definite and the circle is a particular one smcc It then passes through three fixed points, viz., the double point P and the point Q. /, » x In this case ^CQP=^CPO. ' But .iCPQ is given, since P, Q, and the line L are given. ••• ^CQP is given and C is a fixed point. 130'. /V^/;/.7//.-To describe a circle to touch two given non-parallel lines. Let L and M be the lines inter- secting at O. Draw N, N, the bisectors of the angle between Land M.( 121°, Cor. 2) From C, any point on either bi- sector, draw CA ± to L. The circle with centre C and radius CA touches L, and if CB be drawn J. to M, CB = CA. /^gox Therefore the circle also touches M. As C is any point on the bisectors the problem is indefinite, and the centre-locus of a circle which touches two intersecting lines is the two bisectors of the angle between the lines. 131°. /'r^-^/.v/z.-To describe a circle to touch three given lines which form a triangle. L, M, N are the lines forming the triangle. ComfK-DYa^v I„ Ej, the internal and external bisectors of the angle A ; and I^, E.,, those of the angle B. ^ + ^Bis<±, .-. ^BAO + z.ABOis is a square. Then the AABC^ AADC, and they are therefore equal. ® ° Now, if AD and DE be equal and in line, the As ADC and EDC are con- gruent and equal. Therefore the AABC may be taken t from its present position and be put into the position of CDE. And the square ABCD is thus trans- formed into the AACE without any change of area ; nABCD=AACE. It is evident that a plane closed figure may be considered from two points of view. 1. With respect to the character and disposition of the lines which form it. When thus considered, figures group themselves into triangles, squares, circles, etc., where the members of each group, if not of the i;ame form, have at least some community of form and character. 2. With respect to the areas enclosed. When compared from the first point of view, the capability of superposition is expressed by saying that the figures are congruent. When compared from the second point of view, it is expressed by saying that the figures are equal. Therefore congruence is a kind of higher or double equality, that is, an equality in both form and extent of area. This is properly indicated by the triple lines ( = ) for con- gruence, and the double lines (=) for equality. 138°. Def.—ThQ altitude of a figure is the line-segment which measures the distance of the farthest point of a figure from a side taken as base. The terms base and altitude are thus correlative. A tri- angle may have three different bases and as many corre- sponding altitudes. /g-oK In the rectaufrle (9,0°. Dpf '>^ i\vn o/^;o^^„f -iri-e i---* perpendicular to one another, either one may be taken as the COMPARISON OF AREAS. gj base and the adjacent one as the altitude. The rectangle naSS:Af thTSS"^e:L~^ '"'-'^'' ^^ SECTIOxN I. COMPARISON OF AREAS-RECTANGLES PARALLELOGRAMS, TRIANGLES. 2. Equal rectangles with equal bases have equal altitudes. 3. Equal rectangles with equal altitudes have equal bases. I. In the CDS BD and FH, if AD = EH, and AB = EF, then c=iBD=[=iFH. Proo/.~mace E at A and EH along . AD. Then, as z.FEH=^BAD = n, EF will lie along AB. And because EH=AD and EF = AB, therefore H falls at e ual ^^ ''''^ ^^^ ^""^ °' ""'^ congruent and therefore 2. If □BD=(=,FH and AD = EH, then AB = EF. .T^^'^'f ~^^^^ '^ "°^ ^'^"^^ t° A^' let AB be > EF. Make AP = EF and complete the aPD. Then mPn^oFH h- t^^ ^ - , ^ — - >— ^ ") D^ tne iirai part, but oBD=oFH, (,yp.^ B P A F H 94 SYNTHETIC GEOMETRY. tzDl'D-=c=iBD, which is not true, .'. AIJ and EF cannot be unequal, or 3. If i=]I3D=aFIl and AIi = EF, then AD = EH. Proo/.—Let AB and EF be taken as bases and AD and EH as altitudes (138°), and the theorem follows from the second part. ^ ^ ^ Cor. In any rectangle we have the three parts, base, alti- tude, and area. If any two of these are given the third is given also. i;i 140°. Theorem.—^ parallelogram is equal to the rectangle ^ ^ F _c on its base and altitude. AC is a EH] whereof AU is the base and DF is the altitude. Then ZZZ7AC=[=i on AU and DF. Pw^/— Complete the aADFE by drawing AE ± to CD produced. Then AAEB = AUFC, '.• AE = DF, AB = DC, and ^EAB = i.FDC; .-. ADFC may be transferred to the position AEB, and ZZZ7ABCD becomes the oAEFD, ^I=7AC = c=]on AD and DF. q,e.d. Cor. r. Parallelograms with equal bases and equal altitudes are equal. For they are equal to the same rectangle. Cor. 2. Equal parallelograms with equal bases have equal altitudes, and equal parallelograms with equal altitudes have equal bases. Cor. 3. If equal parallelograms be upon the same side of the same base, their sides opposite the common base are in line. 141°. Thcorem.~k triangle is equal to one-half the rect- angle on its base and altitude. COMPARISON OF AREAS. 95 ABC is a triangle of which AC is the base and BE the altitude. g ^ Then AABC= to on AC and BE. ^''^^A— Complete the ^=7ABDC, of which AB and AC are adjacent sides, a Then AABC = AI)CIi, AABC = ^zzZ7AU = An on AC and BE. (140°) g.su/. Cor. I. A triangle is equal to one-half the parallelogram having the same base and altitude. Cor. 2. Triangles with equal bases and equal altitudes are equal. For they are equal to one-half of the same rectangle. Cor. 3. A median of a triangle bisects the area. For the median bisects the base. Cor. 4. Equal triangles with equal bases have equal alti- tudes, and equal triangles with equal altitudes have equal bases. Cor. 5. If equal triangles be upon the same side of the same base, the line through their vertices is parallel to their common base. 142°. T/ic'orem.~U two triangles are upon opposite sides of the same base — 1. When the triangles are equal, the base bisects the seg- ment joining their vertices ; 2. When the base bisects the segment joining their vertices, the triangles are equal. (Converse of i.) b ABC and ADC are two triangles upon opposite sides of the common base AC. 1. If AABC = AADC, ^' then BH = HD. Proo/.—Lot BE and DF be altitudes, Then •.• AABC=AADC, .-. BE = DF, AEBH=AFDH,andBH = HD. 2. IfBH = HD, then AABC=AADC. q.e.d. 96 SYNTHETIC GEOMETRY. Proof.— i^mcc BH=:HD, .-. AAnH=AADH, and ACBH=ACDH. (141°, Cor. 3) .-. adding, AAHC=AADC. g^e.d. 143. /Ay:— By the sum or difference of two closed figures is meant the sum or difference of the areas of the figures. If a rectangle be equal to the sum of two other rectangles its area may be so superimposed upon the others as to cover both. 144°. Theorem.— \{ two rectangles have equal altitudes, their sum is equal to the rectangle on their common altitude and the sum of their bases. B •^ Proo/.—het the cus X and Y, having equal altitudes, be so placed as to have ' E their altitudes in common at CD, and so that one czi may not overlap the other. Then ^HDC=^CDF=n» BDF is a line. (38°, Cor. 2) Similarly ACE is a line. But BU is II to AC, and BA is || to DC || to FE ; therefore AF is the czi on the altitude AB and the sum of the bases AC and CE ; and the [=iAF=[=iAD +OCF. q.e.d. Cor. I. If two triangles have equal altitudes, their sum is equal to the triangle having the same altitude and having a base equal to the sum of the bases of the two triangles. Cor. 2. If two triangles have equal altitudes, their sum is equal to one-half the rectangle on their common altitude and the sum of their bases. Cor. 3. If any number of triangles have equal altitudes, their sum is equal to one-half the rectangle on their common altitude and the sum of their bases. In any of the above, "base" and "altitude" are inter- chanGfcable. COMPARISON OF ARK AS. 97 145 . Theorem. Two lines parallel to the sides of a para lie ogram and intersecting upon a diagonal divide the parallelogram into four parallelograms such that the two through which the diagonal docs not pass are equal to one another. In the ZIi:7ABCD, E F is || to AD and GH is II to BA, and these intersect at O on the diagonal AC. Then Z=I7BO = ziZ70D. /'^^^/-AABC=AADC, and AAEO = AAHO and AOGC=AOFC; (141°, Cor. i) but CZ7B0=AABC-AAE0-A0GC and CZ70D::=AADC-AAH0-A0FC' ^=7BO = £=:70D. ^, . q.e.a. Cor. I. [ZZl^Y = i — iCA^ Cor. 2. If ZZI7BO = £=70U,0 is on the diagonal AC. (Converse of the theorem.) For if O is not on the diagonal, let the diagonal cut EF in O. Then£z:7B0' = zz:^0'D. /,. os But CZ7B0' is < £=:^B0, and CZ70'D is > E=70D ; ■*• ^?^ is >Z=:70D, which is contrary to the hypothesis- .'. the diagonal cuts EF in O. Ex. Let ABCD be a trapezoid. (84°, Def ) In line with AD make DE = BC, and in line with BC make CF = AD. Then BF=AE and BFEA is a OZJ. But the trapezoid CE can be superimposed on the trape- zoid DB, since the sides are respectively equal, and ^F=A, and ;iE = B, etc. trapezoid BD = j^/ — 7 \\v or, a trapezoid is equal to ore hdf the rectanale on its alti- tude and the sum of its bases. G i«'i< H i. ; % 1 '! 1 98 SYNTHETIC GEOMETRY. I I' Exercises. 1. To construct a triangle equal to a given quadrangle. 2. To construct a triangle equal to a given polygon. 3. To bisect a triangle by a line drawn through a given point in one of the sides. 4. To construct a rhombus equal to a given parallelogram, and with one of the sides of the parallelogram as its side. 5. The three connectors of the middle points of the sides of a triangle divide the triangle into four equal triangles. 6 Any line concurrent with the diagonals of a parallelogram bisects the parallelogram. 7. The triangle having one of the non-parallel sides of a trapezoid as base and the middle point of the opposite side as vertex is one-half the trapezoid. 8. The connector of the middle points of the diagonals of a quadrangle is concurrent with the connectors of the middle points of opposite sides. 9. ABCD is a parallelogram and O is a point within. Then AAOB + ACOD = A/ — 7. What does this become when O is without ? TO. ABCD is a parallelogram and O is a point within. Then AAOC = AAOD-AAOB. What does this become when O is without.? (This theorem is important in the theory of Statics.) Bisect a trapezoid by a line through the middle point of one of the parallel sides. By a line through the middle point of one of the non-parallel sides. The triangle having the three medians of another tri- angle as its sides has three-fourths the area of the other. TI 12 COMPARISON OF ARliAS. POLYGON AND CIRCLE. 99 i i 146°. I},f.~The sum of all the sides of a polvgon is called .^A..w/,-. and when the polygon is regular 'every sid is a. the same d.stance from the centre. This distance is the apothem of the polygon. Thus if ABCD...LA be a regu- lar polygon and O the centre (132", l^Qi. 2), the triangles OAH, OBC, ... are all congruent, and OP = OQ=:etc. AB + BC + CD + ... + LA is the perimeter a^nd OP, per- pendicular upon AB, is the apothem. ^ 147°. Theorem.~A regular polygon is equal to one-half the rectangle on its apothem and perimeter. Jrf'~l^^ '"'T^^^' ^^^' ^°^' - LOA have equal altitudes, the apothem OP, .-. their sum is one-half the o on 01 and the sum of their bases AB-h BC + ... LA. (144°, Cor i) But the sum of the triangles is the polygon, and the sum o their bases is the perimeter. .-. a regular polygon =|a on its apothem and perimeter. • \f' ^{"^ ^'"''^-^ i^^nit or limiting value of a variable IS the value to which the variable by its variation can be made to approach indefinitely near, but which it can never be made to pass. Let ABCD be a square in its cir- cumcircle. If we bisect the arcs AB, BC, CD, and DA in E, F, G, and H,' we have the vertices of a regular octagon AEBFCGDHA. Now, the area of the octagon appmaches nearer to that of the circle than the area of the square does ; and the m ■E ! imk l^H ii msm 11 r %'i. ■ II \\ fl 11 •n H H ,'3 l^^l 100 SYNTl 1 KTIC GL:uM KTRY. 1 li perimeter of the octagon approaches nearer to the Icn.^th of the circle than the perimeter of the sciuare does ; and the apothem of the octagon approaches nearer to the radius of the circle than the apothem of the square does. Again, bisecting the arcs AK, KH, BK, etc., in I, J, K, etc., we obtain the regular jjolygon of i6 sitles. And all the fore- going parts of the polygon of i6 sides approach nearer to the corresponding parts of the circle than those of the octagon do. It is evident that by continually bisecting the arcs, we may obtain a series of regular polygons, of which the last one may be made to approach the circle as near as we please, but that however far this process is carried the final polygon can never become greater than the circle, nor can the final apothem become greater than the radius. Hence the circle is the limit of the perimeter of the regular polygon when the number of its sides is endlessly increased, and the area of the circle is the limit of the area of the poly- gon, and the radius of the circle is the limit of the apothem of the polygon under the same circumstances. 149°. Theorem. —Ps. circle is equal to one-half the rectangle on its radius and a line-segment equal in length to the circle. Proof. — The is the limit of a regular polygon when the number of its sides is endlessly increased, and the radius of the is the limit of the apothem of the polygon. But, whatever be the number of its sides, a regular polygon is equal to one-half the cz] on its apothem and perimeter. (147"") .*. a is equal to one-half the o on its radius and a line- segment equal to its circumference. EXKRCISKS. I. Show that a regular polygon may be described about a circle, and that the limit of its perimeter when the number of its sides is increased indefinitely is the circumference of the circle. MKASUKEMENT ()K I.KNCITHS AND ARKAS. lOI 2. The difference between the areas of two regular polygons, one inscribed in a circle and the other circumscribed about it, vanishes at the limit when the number of sides of the polygons increases indefinitely. 3. What is the limit of the internal angle of a reguhir polygon as the number of its sides is endlessly increased ? SECTION II. MEASUREMENT OF LENGTHS AND AREAS. 150". Ih'f.—\. That part of Geometry which deals with the measures and measuring of magnitudes is Mctriuxl Geometry. 2. To measure a magnitude is to determine how many unit magnitudes of the same kind must be taken together to form the given magnitude. And the number thus determined is called the measure of the given magnitude with reference to the unit cmi)loyed. This number may be a whole or a frac- tional number, or a numerical quantity which is not arith- metically expressible. The word "number" will mean any of these. 3. In measuring length, such as that of a line-segment, the unit is a segment of arbitrary length called the imit-lcHgUi. In practical work we have several such units as an inch, a foot, a mile, a metre, etc., but in the Science of Geometry the unit-length is quite arbitrary, and results obtained through it are so expressed as to be independent of the length of the particular unit employed. 4. In measuring areas the unit magnitude is the area of the square having the unit-length as its side. This area is the unit-area. Hence the unit-length and unit-area are not both 102 SYiNTllETlC GEOMETRY. arbitrary, for if either is fixed the other is fixed also, and determinable. This relation between the unit-length and the unit-area is conventional, for we might assume the unit-area to be the area of any figure which is wholly determined by a single segment taken as the unit-length : as, for example, an equi- lateral triangle with the unit-length as side, a circle with tiie unit-length as diameter, etc. The square is chosen because it offers decided advantages over every other figure. For the sake of conciseness we shall symbolize the term unit-length by u.L and unit-area by u.a. 5. When two magnitudes are such that they are both capable of being expressed arithmetically in terms of some common unit they are commensunxblc, and when this is not the case they are incomfuensurahlc. E B P ' lilus.—hQt ABCD be a sc|uare, and let EF and HG be drawn _L to BU, and EH and FG ± to AC. Then EFGH is a square (82°, Cor. 5), and the triangles AEB, APB, BFC, BPC, etc., are all equal to one another. If AB be taken as it./., the area of the square AC is the u.a.'y and if EF be taken as u.L, the area of the square KG is the 71 a. In the first case the measure of the square AC is i, and that of EC; is 2 ; and in the latter case the measure of the square EG is r, and that of AC is h So that in both cases the measure of the square EG is double that of the square AC. .'. the squares EG and AC are commensurable. Now, if AB be taken as u.L, EF is not expressible arith- metically, as will be shown hereafter. .'. AB and EF are incommensurable. p H D G 151°. Let AB be a segment trisected at E and F (127°), and let AC be the square on AB. Then AD=AB. And M 1 2 3 4 ^ 6 7 8 9 H MEASUREMENT OF LENGTHS AND AREAS. I03 if AD be trisected in the points K and M, and through E and F ||s be drawn to AD, and through K and M ||s be drawn to Ali, the figures '^ — ^^ ^ '> -> 3> 4> 5> 6, 7, 8, 9 are all squares equal to k one another. Now, if AH be taken as «./., AC is the «.r?. ; and if AE be taken as «./., any one of the small squares, as AP, is the um. And the segment AB con- tarns AE 3 times, while the square AC contains the square AP m three rows with three in each row, or 3-' times. .'. if any assumed ?i./. be divided into 3 equal parts for a new «./., the corresponding u.a. is divided into f equal parts for a new u.a. And the least consideration will show that this is true for any whole number as well as 3. .-. I. If an assumed u./. be divided into ;/ equal parts for a new u./., the corresponding ti.a. is divided into n^ equal parts for a new u.a.; n denoting any whole number. Again, if any segment be measured by the u.l. AB, and also by the u.l. AE, the measure of the segment in the latter case is three times that in the former case. And if any area be measured by the u.a. AC, and also by the u.a. AP, the measure of the area in the latter case is f times its measure m the former case. And as the same relations are evidently true for any whole number as well as 3, .-. 2. If any segment be measured by an assumed ti.l. and also by ^^th of the assumed //./. as a new 71.I., the measure of the segment in the latter case is ;/ times its measure in the former. And if any area be measured by the corresponding u.a.^ the measure of the area in the latter case is n^ times its measure in the former case ; ;/ being any whole number. This may be stated otherwise as follows :— By reducing an assumed u.l. to ^h of its original length, we increase the measure of any given segment ;/ times, and we mcrease the measure of any given area n^ times ; n beino- a whole number. ° * t^H ^^H (^^1 1^1 «« i HI^^H '^. 104 SYNTHETIC GEOMETKY. In all cases where a //./. and a u.a. are considered together they are supposed to be connected by the relation of i ;o°' 3 and 4. •' ,' 152'. 7y/6'^m//.-The number of unit-areas in a rectangle IS the product of the numbers of unit-lengths in two adjacent sides. The proof is divided into three cases. I. Let the measures of the adjacent sides with respect to ^ c the unit adopted be whole numbers. Let AB contain the assumed //./. a times, and let AD contain it d times. ^ Then, by dividing AB into a equal D parts and drawing, through each point of division, lines || to AD, and by dividing AD into b equal parts and drawing, through each point of division, lines || to AB, we divide the whole rectangle into equal squares, of which there are a rows with /; squares in each row. the whole number of squares is ab. But each square has the nJ. as its side and is therefore the lui. ''•'^-s in AC^7/./.s in AB x u.l.s in AD. We express this relation more concisely by writing symbolic- '% i=iAC = AB.AD, where nAC means " the number of u.a.s in cnAC," and A B and AD mean respectively "the numbers of «./.s in these sides." And in language we say, the area of a rectangle is the pro- duct of Its adjacent sides ; the proper interpretation of which is easily given. 2. Let the measures of the adjacent sides with re?pect to the unit adopted be fractional. Then, •.• AB and AD are commensurable, some m'^i will be an aliquot part of each (150°, 5). Let the new unit be -ih. of the adopted unit, and let AB contain p of the new units, and AD contain g of them. The measure of oAC in terms of the new n.a. is Pq MEASUREMKNT OF LENGTHS AND AREAS. 105 ( 1 52^ I ), and the measure of the oAC in terms of the adopted unit is pq «5 „ - (151°, 2) But the measure of AH in terms of the adopted ,../ is ^, and of AD it is y. ^ and ;/ or li^ n ' n aAC=:AIJ.AD. ////,,-. Suppose the measures of Ali and AD to some un,t-.en AD ; .-. AE = AD aAC=AB.AD. or q.e.d. m\ i't \ io6 SYNTHETIC GEOMETRY. IHr !■ 1 1 1 ^^■(■H IH I^H 19 ^^B ^ H I S3''- The results of the last article in conjunction with Section I. of this Part give us the following theorems. 1. The area of a parallelogram is the product of its base and altitude. /,.^o\ (140) 2. The area of a triangle is one-half the product of its base and altitude. /,..o\ 3. The area of a trapezoid is one-half the product of its altitude and the sum of its parallel sides. (145°, Ex.) 4- The area of any regular polygon is one-half the product of its apothem and perimeter. (147°) 5. The area of a circle is one-half the product of its radius and a line-segment equal to its circumference. (149°) Ex. I. Let O, O' be the centres of the in-circle and of the ex-circle to the side BC (13:°); and let OD, O'P" be perpen- diculars on BC, OE, O'P' per- pendiculars on AC, and OF, O'P on AB. Then ^^ , OD = OE = OF = ?' ^ E c\ P ^,^^^ 0'P = 0'P'=0'P''=r'; .-. AABC=AAOB-|-ABOC + ACOA = UB.OF + U)C.OD + ^CA.OE (153,2) = V,;- X perimeter = rs, where j is the half perimeter ; Ex.2. AABC = AAO'B + AAO'C-ABO'C =iO'P . AB + iO'P'. AC - iO'P". BC where r' is the radius of the ex-cirrle to side a ; A=r'(s~(7). Similarly, ^ = r"(s-d) = r"'{s-c). MEASUREMENT OF LENGTHS AND AREAS. 10/ I. '=.l + ^ +± r r' r" r'" .''.lit Exercises. 2. /^i=rr'r"r' 3. What relation holds between the radius of the in-circle and that of an ex-circle when the triangle is equiangular .> A^/..-When the diameter of a circle is taken as the u I the measure of the circumference is the inexpressible numeri- cal quantity symbolized by the letter ir, and which, expressed approximately, is 3. 141 5926.... 4. What is the area of a square when its diagonal is taken as the ti.l. ? 5. What is the measure of the diagonal of a square when the side is taken as the u.U /j -^o . 6. Find the measure of the area of a circle when the di- ameter is the n.L When the circumference is the tc l. 7. If one line-segment be twice as long as another, the square on the first has four times the area of the square on the second. (151° 2) 8. If one line-segment be twice as long as another, 'the equilateral triangle on the first is four times that on the second. . 9. The equilateral triangle on the altitude of another equitat- eral triangle has an area three-fourths that of the other 10. The three medians of any triangle divide its area into SIX equal triangles. From the centroid of a triangle draw three lines to the sides so as to divide the triangle into three equal quadrangles. In the triangle ARC X is taken in BC, Y in CA and Z in AB, so that BX = ^BC, CY=]CA, and AzL^ak Express the area of the triangle X YZ in terms of that of ABC. 1 [ 12 13. Generalize 12 by making BX='bC etc n ' 14. Show that a-si\ - ^\=.L{r" -^-r'"). io8 SYNTHETIC GEOMETRY. P SECTION III. GEOMETRIC INTERPRETATION OF ALGEBRAIC FORMS. 1 54°. We have a language of symbols by which to express and develop mathematical relations, namely, Algebra. The symbols of Algebra are quantitative and operative, and it is very desirable, while giving a geometric meaning to the symbol of quantity, to so modify the meanings of the sym- bols of operation as to apply algebraic forms in Geometry. This application shortens and generalizes the statements of geometric relations without interfering with their accuracy. Elementary Algebra being generalized Arithmetic, its quantitative symbols denote numbers and its operative sym- bols are so defined as to be consistent with the common properties of numbers. Thus, because 2 + 3 = 3-1-2 and 2.3 = 3.2, we say that (i->(-b = b + a and ab = ha. This is called the commutative law. The first example is of the existence of the law in addition, and the second of its existence in multiplication. The commutative law in addition may be thus expressed : — A sum is independent of the order of its addends ; and in multiplication — A product is independent of the order of its factors. Again, because 2(3 4-4) = 2 . 3 + 2 . 4, we say that a{b + c) = ab + ac. This is called the distributive law and may be stated thus : — The product of multiplying a factor by the sum of several terms is equal to the sum of the products arising from multiplying the factor by each of the terms. These two are the only laws which need be here mentioned. And any science vvhich is to employ the forms of Algebra INTERPRETATION OF ALGEBRAIC FORMS. 109 must have that, whatever it may be, which is denoted by the algebraic symbol of quantity, subject to these laws. 155°. As already explained in 22° we denote a single line- segment, m the one-letter notation, by a single letter, as a which IS equivalent to the algebraic symbol of quantity ; and A single algebraic symbol of quantity is to be interpreted geometrically as a line-segment. It must of course be understood, in all cases, that in em- ploying the two-letter notation for a segment (22°) as " AB " the two letters standing for a single line-segment are equiva- lent to but a single algebraic symbol of quantity. The expression a + d denotes a segment equal in length to those denoted by a and b together. Similarly 2a=a + a, and na means a segment as long as n of the segments a placed together in line, n being anv numerical quantity whatever. r^^^. a-b, when a is longer than b, is the segment which is left when a segment equal to b is taken from a. Now it is manifest that, if a and b denote two segments a^bj^b^a, and hence that the commutative law for addition applies to these symbols when they denote magnitudes having length only, as well as when they denote numbers. IS6^ Lme in Opposite Senses.- A quantitative symbol a is in Algebra always affected with one of two signs -f- Jr - which, while leaving the absolute value of the symbol un- changed, impart to it certain properties exactly opposite in character. This oppositeness of character finds its complete interpreta- tion in Geometry in the opposite directions of every segment Thus the segment in the margin may be con- a sidered as extending/r^w A to B or from B to A. A S With the two-letter notation the direction can be denoted by the order of the letters, and this is one of the advantages HO SYNTHETIC GEOMETRV. of this notation; but with the one-letter notation, if we denote the segment AB by +a, we must denote the segment BA by -^. But as there is no absolute reason why one direction rather than the other should be considered positive, we express the matter by saying that AB and BA, or +^and -«, denote the same segment taken in opposite semises. Hence the algebraic distinction of positive and negative as applied to a single symbol of quantity is to be interpreted geometrically by the oppositeness of direction of the segment denoted by the symbol. Usually the applications of this principle in Geometry are confined to those cases in which the segments compared as to sign are parts of one and the same line or are parallel. Ex. I. Let ABC be any A and let BD be the altitude from the vertex B. Now, suppose that the sides AB and BC undergo a gradual change, so that B may move along the line BB' until it comes into the position denoted by B'. Then the segment AD gradually di- minishes as D approaches A ; disappears when D coincides with A, in which case B comes to be vertically over A and the A becomes right-angled at A ; reappears as D passes to the left of A, until finally we may suppose that one stage of the change is represented by the AAB'C with its altitude B'D'. Then, if we call AD positive, we must call AD' negative, or we must consider AD and AD' as having opposite senses.' Again, from the principle of continuity (104°) the foot of the altitude cannot pass from D on the right of A to D' on the left of A without passing through every intermediate point, and therefore passing throug/i A. And thus the seg- ment AD must vanish before it changes sign. This is conveniently expressed by saying that a line- INTERPRETATION OF ALGEBRAIC FORMS. 1 1 1 segment changes sign 'when it passes through zero ■ nassin. through zero being interoreted a^ vnnJc^- '-^^ ^^^""^ passing on the other side of th'Topoim '"^^ '"' "^^^^^^'"^ Ex. 2^ ABCD is a normal quadrangle. Consider the side AD and suppose D to move along the line UA until It comes into the position D' ^ The segments AD and AD' are opposite m sense, and ABCD' is a crossed quad- rangle. ^ .-. the crossed quadrangle is derived from the normal one by changing the sense of one of the sides c lit, -ire ceases to be a crossed quadrangle. alld'bu; Jl!!ll '' "'' ''"""P^' ^^''" ''^^''''' ^'hich are par- allel but which are not in line have opposite senses. ABC is a A and P is any point within from which perpendiculars PD, PE, PF are drawn to the sides. Suppose that P moves to P'. Then PF becomes P'F', and PF and P'F' being in the same direc- tion have the same sense. Similarly PE becomes P'E' ^d these segments have the same sense. But PD becomes P D which IS read in a direction opposite to that of PD Hence PD and P'D' are opposite in sense. But PD and P'D' are perpendiculars to the same line from points upon opposite sides of it, and it is readily seen that in passing from P to P' the ±PD becomes zero and'then chang sense as P crosses the side BC. Hence if by any continuous change in a figure a point passes from one side of a line to the other side' the pe.p n dicular from that point to the line changes sense. Cor. If ABC be equilateral it is easily shown that PD + PEH-PF = a constant. m • if m I T2 SVNTIIl-TIC GEOMETRY And if we regard the sense of the segments this statement is true for all positions of P in the plane. 157°. Product ~Y\,^ algebraic form of a product of two symbols of quantity is interpreted geometrically by the rect- angle having for adjacent sides the segments denoted by the quantitative symbols. This is manifest from Art. 152°, for in the form ab the smg e letters may stand for the measures of the sides, and the product ab will then be the measure of the area of the rect angle. If we consider ab as denoting a □ having a as altitude and b as base, then ba will denote the o having b as altitude and a as base. But in any □ it is immaterial which side is taken as base (138 ) ; therefore ab=ba, and the form satisfies the commutative law for multiplication. Again, let AC be the segment b^-c, and AB be the segment - ° ^ 'h so placed as to form the n2a{b^-c) or AF. Taking AD =/;, let UE be drawn II to AB. Then AE and DF are rect- angles and DE=AB=rt. oAE is ^ab, and oDF is zimc ; ^=^n{b^-c)=U2ab^r^:nac^ and the distributive law is satisfied. 158°. We have then the two following interpretations to which the laws of operation of numbers apply whenever such operations are interpretable. I. A smgle symbol of qtiantity denotes a line-se^ment As the sum or difference of two line-segments is a segment the sum of any number of segments taken in either sense is a segment. Therefore any number of single symbols of quantity con- nected by + and - signs denotes a segment, as a-^b '7 -b + e, a- b + (-c\ etc. ' INTER]»RETATION OF AUIKKRAIC FORMS. 1 13 For this reason such expressions or forms are often called linear^ even in Algebra. Other forms of linear expressions will appear hereafter. 2. The product form of two symbols of quantity denotes the rectangle whose adjacent sides are the sci^ments denoted by the single symbols. A rectangle encloses a portion of the plane and admits of measures m two directions perpendicular to one another, hence the area of a rectangle is said to be of two dimensions And as all areas can be expressed as rectangles, areas in general are of two dimensions. Hence algebraic terms which denote rectangles, such as ab, {a-^b)c, {a + bXc-^d), etc., are often called rectangular terms and are said to be of two dimensions. ad c ' Ex. Take the algebraic identity a{b+c) = ab-{-ac. The geometric interpretation gives— If there be any three segments {a, b, c) the c=] on the first and the sum of the other two (/., c) is equal to the sum of the as on the first and each of the other two. The truth of this geometric theorem is evident from an mspection of a proper figure. This is substantially Ejiclid, Book II., Prop. i. 1 59°- Square.— When the segment b is equal to the seg- ment a the rectangle becomes the square on a. When this equality of symbols takes place in Algebra we write a^ for aa and we call the result the " square » of a, the term " square "' being derived from Geometry. Hence the algebraic form of a square is interpreted geo- metrically by the square which has for its side the segment denoted by the root symbol. Fa'. In the preceding example let /; become equal to a, and a{a + c) = a--\-ac, H m m. 1J4 SViNTIlKTIC GliOMKTRV. which interpreted ^geometrically pives— If a segment (u + c) be divided into two parts (rt, r), the rectangle on the segment and one of its parts (u) is equal to the sum of the square on that part (a-) and the rectangle on the two parts { + c is not interpretable geometrically This is expressed by saying that— An alge- braic form has no geometric interpretation unless the form is Jwviogcncous, i.e., unless each of its terms denotes a geo- metric element of the same kind. It will be observed that the terms "square," "dimensions," "homogeneous," and some others have been introduced into Algebra from Geometry. 161°. Rectaugies i?t Opposite Senses. —The algebraic term alf changes sign if one of its factors changes sign. And to be consistent we must hold that a rectangle changes sense whenever one of its adjacent sides changes sense. Thus the rectangles AB . CD and AB . DC are the same in extent of area, but have opposite senses. And AB.CD + AB.DC = o, fo^ thesum = AB(CD-f DC), and CD + DC = o. (156°) INTKRPRKTATION OF ALGEin''" '^ !>PP'»* • •• AB . OF and A'B . O'F' have the same senlf ' ^'' ^^ Qd. A'BtD=ADO'C- AA-O'B ; ilH Ii6 SVNTHICTIC (iKOMKTRY. or, the area of a crossed quadrangle must be taken to l)e the difference between the two triangles which constitute it. 162°. Theorem. ~K quadrangle is equal to one-half the parallelogram on its diagonals talien in both magnitude and relative direction. AnCD is a (juadrangle of which AC and liD are diagonals. Through H and 1) let P(2and RS be drawn || to AC, and through A and C let PS and ()R be drawn || to IJD. Then P()RS ]s the ZZZ7 on the diagonals AC and BD in both magnitude and direction. Qd. ABCD-AzZZ7P()RS. Proof.— Ci^i. A PCD -AAPC+AAI)C(istFig.) = AABC - AADC (2ndFig.) (161°, Kx.) Aabc^Aezvpoca, AAI)C = i£Z:7SRCA, (141°, Cor. i) Qd. ABCD-^zz::^P()RS in both figures. This theorem illustrates the generality of geometric results when the principle of continuity is observed, and segments and rectangles are considered with regard to sense. Thus the principle of continuity shows that the crossed cfuadrangle is derived from the normal one (156°, Ex. 2) by changing the sense of one of the sides. This requires us to give a certain interpretation to the area of a crossed quadrangle (161°, Ex. i), and thence the present example shows us that all quadrangles admit of a common expression for their areas. 163°. A rectangle is constructed upon two segments which are independent of one another in both length and sense. But a square is constructed upon a single segment, by using ( ' > lNTKkl>RKTATION OF ALGKIiKAIC KORMS. I 17 it for each side. In other words, a rectangle depends upon two segments while a square depends upon only one Hence a scjuare can have only one sign, and this is the one which we agree to call positive. Hence n square is always positive. 164^ The algebraic equation ab^cd tells us gepmetrically that the rectangle on the segments a and /; is equal to the rectangle on the segments c and d. Hut the same relation is expressed algebraically by the form cd a — b' therefore, since a is a segment, the form "L is linear and denotes that segment which with a determines a rectangle equal to cd. Hence an expression such as ''^' j^^'^' ^^^-f j^ Xxx\^,xx c a b 165°. The expression a-^bc tells us geometrically that the square whose side is a is equal to the rectangle on the seir- ments b and c. But this may be changed to the form a = s'br. Therefore since .^ is a segment, the side of the square, the form >Jbi- is linear. Hence the algebraic fonn of the square root of the product of two symbols of quantity is interpreted oca metrically by the side of the square which is equal to the rectaiii^le on the segments denoted by the quantitative symbols. 166°. The following theorems are but geometric interpreta- tions of well-known algebraic identities. They mav, however be all proved most readily by superposition of areas,' and thus the algebraic identity may be derived from the o-eo- metric theorem. *' ,P^ 4 1 1 1 i If 1^ us SVNrilKTIC (IKOMiaKV. I. The .square on the sum of two sckmiiciUs is equal to ll»e sum of ihc squares on the si>Kmcnts and twice the '' rectangle on the scKUjents. (,M /O" -«/■' + //^ + 2y and EG is " ffl4b)' ^ {)( + r) + l>c. 4 State and prove geometrically by superposition of areas ab b ab h b ab E F (n-b)^ L Q b ab c. AKI:AI. klOLATIONS. 119 denote sci^nicnts. 5. If a Kivcn scKmcnt l,c (livi.U.l into any three parts the square on the segment is ecpial to the sum of the squares on the parts together with twice the sum of the rectangles on the parts taken two and two. ••"7' ^^>' <^«'"Pa>ison of areas from the Fig. of Ex. ,, ,ha, '^ + /'>-2/>U + fi)^2/Ku-fi)+(a.-/.f, and state the tlicorem in words. SECTION IV. AKKAl. KKI.ATIONS. 167". A/. I. The segment which joins two given points •s called the >/. of the points ; and where no 'eferenc: ! made to ength the ,/./,, of two points may be taken to mean the hne determined by the points. 2. The foot of the perpendicular from a given point to a given hne .s the .;-//..,.,.,/ ^,,y,r//.;,, or simply L projcc tton, of the pomt upon the line. 3. length being considered, the join of the projection of two pomts IS the projection of the Q join of the points. Thus if L be a given line and P, Q, two given points, and PI'', QQ' perpendiculars upon L ; PQ is the join of P and (), P' and Q' are the p q— t projections of P and (2 upon L, and the segment P'O' is fh. projection of PQ upon L. ^ ^ "^^ 16S . Theorem. -The sum of the projections of the sides of 'r i' 1 m \>:' VV^ ^ I 1^1 {^^1 \\ \ 1 l^^^l ij^i M^^^^ .) -1 ^^^H 120 SYNTHETIC GEOMETRY any closed rectilinear figure, taken in cyclic order with respect to any line, is zero. A BCD is a closed rectilinear figure and L is any line. Then Pr.AB + Pr.BC + Pr.CD + Pr.DA=o A' B^^ ~^' D' L Proof. ~T>r2i\\\.hQ perpendiculars AA', BB', CC, DD', and the sum of the projections becomes A'B +B'C' + C'D'4-D'A'. But D'A' is equal in length to the sum of the three others and is opposite in sense. .-. the sum is zero. It is readily seen that since we return in every case to the point from which we start the theorem is true whatever be the number or disposition of the sides. This theorem is of great importance in many investigations. Cor. Any side of a closed rectilinear figure is equal to the sum of the projections of the remaining sides, taken in cyclic order, upon the line of that side. Def.—ln a right-angled triangle the side opposite the right angle is called the hypothenuse, as distinguished from the remaining two sides. 169°. Theorem.~\w any right-angled triangle the square on one of the sides is equal to the rectangle on the hypothenuse and the projection of that side on the hypothenuse. ABC is right-angled at B, and BD is ± AC. Then AB2 = AC.AD. Proof.— "L^t AF be the Q on AC, and let EH be II to AB, and AGHB be a □, since L& is a "J. ^GAB = /_EAC = -I, (82°, Cor. 5) z.CAB = ^EAG. AE = AC, (hyp.) ACAB^AEAG, (64°) AG = AB, and AH is the □ on AB. Then Also, ana AREAL KKLATIONS. 121 Now t.e.. (82^ Cor 5) (52^) (I4i"j nAH=.^=7ABLE = aADKE, (140^ AB'^ = AC.AD. \J^ As this theorem is very important we give an alternative proof of It. Proo/.~AT is the □ on AC and AH IS the n on AB, and BD is ± AC "•• ^C'AB=^CAE=n. (82°, Cor. 5)G ^GAC=z.BAE. Also, AG=AB, and AC = AE, .-. AGAC-ABAE. But AGAC = |nAH, and ABAE = ^[=iAK, nAH=nAK, i.e., AB^=AC.AD. Cor. I. Since AB'^= AC . AD we have from symmetry BC- = AC.DC, •*. adding, AB2+ BC-' = AC(AD + DC) or AB^+BC2=AC2. .;. The square on the hypothenuse of a right-angled triangle is equal to the sum of the squares on the reniumng sides This theorem, which is one of the most important in the whole of Geometry, is said to have been discovered by rythagoras about 540 b.c. Cor. 2 Denote the sides by a and . and the hypothenuse ^yb and let a, and c, denote the projections of the sides a and c upon the hypothenuse. ^"^ d'^c^=b\ Cor, 3. Denote the altitude to the hypothenuse by f, Then b = c,^a,, and ADB and CDB are right-angled at D add 2/2 to each side and Uoo , i) b'+2f=.ci^^P^^a{^^.p'^^2c,a,, \\ . i : rif] - il8 12'> SYNTHETIC GEOiMETRV. or c' + a'^-^2p-^ = c"+a^^.2c,a,. (Cor. i) or BI>' = AD.DC, i.e., the square on the altitude to the h> pothenuse is equal to the rectangle on the projections of the sides on the hypo- thcnuse. ^€/-— 'I'hc side of the square ecjual in area to a given rectangle is called the mean proportional or the geometric mean between the sides of the rectangle. Thus the altitude to the hypothenuse of a right-angled A is a geometric mean between the segments into which the altitude divides the hypothenuse. (169", Cor. 3) And any side of the A is a geometric mean between the hypothenuse and its projection on the hypothenuse. (169") 170°. Theorem.— \{ the square on one side of a triangle is equal to the sum of the squares on the remaining sides, the triangle is right-angled at that vertex which is opposite the side having the greatest square. (Converse of 169°, Cor.) If AC^ = ABH BC-', the .iB is a -J- 7';'^^/- Let ADC be a |0 on AC. AC=^ = ABHBC2, ABis ; 17 13 20 1 24 ! 35 48 53 63 68 80 99 ( 1 i ' 29 40 85 ro4 - 1 1 -: I u 20 24 28 32 36 40 1 5 12 : t 21 1 32 45 60 77 _96 109 • • • ■25 "3 .-. . i 34 45 58 7?, 90 J 24 1 1 \ mm 30 , 3b 42 48 54 60 \ \ 7 16 27 j 40 55 72 91 • • • 4 41 ! 52 I 65 80 97 ri6 9 ' 48 56 64 72 80 . 40 i 20 33 48 65 84! ... s 1 i 1 61 60 74 70 89 80 106 90 I 125 ... ICO ... ri 24 1 39 561 75 • • • • • • 1 ... ... • • • f] • ^ A 124 SYNTHKTIC GEOMKTRY. 172. Let 11, h, c be the sides of any triangle, and let b be taken as base. Denote the projections of a and c on /; by a^ and c\, and the altitude to /M)y/. Then (i) //-•^,-,-' + ,^,2 4.2,y,j, (,66°^ ,) (2) Cx'^P\ (169°, Cor. I (3) ^^'' = ih'-\-pK I. IJy subtracting; (2) from (3) .'. The difference behoeen the squares upon hvo sides of a iriangie is equal to the difference of the squares on the projec- tions 0/ these sides on the third side, taken in the same order. Since all the terms are squares and cannot change sign (163°), the theorem is true without any variation for all As. 2. 13y adding (i) and (2) and subtracting (3), b'^ + c^-d' = 2c^' + 2c,a^ -■=2k\, '.' b = Ci + ^i, d^^b'^ + c'^-2k\. Now, since we have assumed that b^e^ + ay, where c^ and a^ are both positive, D falls between A and C, and the angle A is acute. .•. /// an_y triano/e the square on a side opposite an acute ano/e is /ess than the sum of the squares upon the other two sides by twice the rectangle on one of these sides and the pro- jection of the other side upon it. 3. Let the angle A become obtuse. Then D, the foot of the altitude to b, passes beyond A, and c\ changes sign. .'. im/^rj^ changes sign, (161") and d^=b'-\-c--\-2bc.^. .-. The square on the side opposite the obtuse anole in an obtuse- angled triangle is greater than the sum of the squares on the other two sides by twice the rectangle on one of these sides and the projection of the other side upon it. ARKAI. RELATIONS. 125 The results of 2 and 3 are fundamental in the theory of tnanjrjes. ^ These results are but one ; for, assuming as we have done that the rnbc, ,s to be subtracted from Ifi^c^ when A is an acute an- e, the change in sign follows necessarily when A becomes obtuse, since in that case the □ changes sign because one of Its sides changes sign (161°); and in conformity to algebraic forms - ( - ibc^== + 2bcy Cor. If the sides a, b, c of a triangle be given in numbers, we have from 2 c — ^^" + ^""- d" ^ 2b ~~' which gives the projection of c on b. If^i is + the^A is acute; if Cy is o the z.A is ~\ ; ^'^"^ if^iis - the Z.A is obtuse. Ex. The sides of a triangle being 12, 13, and 4, to find the character of the angle opposite side 13. Let ii = a, and denote the other sides as you please, e.cr ^=12 and r= 4. Then ' '^ ' = 1^1+4^3=^^ _ 3 24 8' and the angle opposite side 13 is obtuse. 173;. 7y/.mm-The sum of the squares on any two sides of a tnang e is equal to twice the sum of the squares on one- half the third side and on the median to that side. BE is the median to AC. Then AB^+lJC2=2(AE-'+EB^). Proof.~\.(ti D be the foot of the altitude ^ '^ ^ c on AC. Consider the AABE obtuse-angled at E and AB^' = AE-'-}-EBH2AE.ED. ' (170° ,) Next, consider the ACBE acute-angled at E and BC^-EC2 + EB2-2EC.KD. (17.° .) H.-il •w Ml 126 SYNTHETIC (JEOMETRV. Now, addin^r and rcmcmbciin«,r that AE = EC q.c.d. Cor. I. Denoting the median by ;// and the side upon which It falls by b, wc have for the length of the median 4 Cor. 2. All the sides of an equilateral triangle are equal and the median is the altitude to the base and the ri Z ? u .'^""^^' """' '"' eounllfv .1, ' •'""' "'"^ """I' Siides of the Ztlfr' '"" '°^'="^" "^ "^^^ P-^ '".ough .ore by Now, of the two segments from B we always know which .s the greater by 63°, and if we write PA for Al' the oPA PC "::*"'?■"" ''r*^^^''-^^- He„ce;co°:it„' theoJem-' "*■' "'"'^^ P"^'"™' "« '"^'X ^'^o the 175''- I. From 174° we have I3A2-BP2 = AP VC m« BA .s fixed, therefore the oAP.PC increats'as BP 7 creases. But BP is least when P is at d7^^ ^\^ the CAP . PC is greatest when P is at D. ^ ' ' ^' ''""''" Def. i.-A variable magnitude, which by continuous change may „.crease until a greatest value is reac ed an t" esT" 1"' '^ ^f ^^ '^ ^--^P'^^^'^ ^' ^ -aximm^ td X greatest value reached is its maximum Thus as P moves from A to C the oAP.PC increases from zero, when P is at A, to its maximum value, v en P ^^ And as AC may be considered to be any segment divided .;. Themaxhnum rectangle ou the parts of a give7i se^^^{a-¥b + c){b+c-a){c-^a-b){a + b-c\ and by writing s for ^(^- + ^ + 4 and accordingly s-a for \{b+c-a), etc., we ob tain A=sl7{J^){7^b)(^cY. This important relation gives the area of the A in terms of its three sides. Ex. 2. Let ABC be an equilateral A- Then the area may be found from Ex. i by making a = b = c, when the reduced expression becomes, A^'^\'3- AKEAL RELATIONS. joQ circ^ idt'"' ^'^ "" ^'^ '■^^"^^^' °^^'^^- - 7"- of its Let A B, C be three vertices of the octagon and O the centre. Complete the square OD. and draw BE ± to OA. Since^EOB-J-]. andOB = ;' EO=EB==|;V2, and AOAB = iOA.EB = J^r.irV2=l;V2 But AOAB is one-eighth of the Jctagon Oct. = 2rV2. « > Exercises. .. ABC is right-angled at B, and E and K are middle points , °fBA and lie respectively. Then 5AC^=4(CEHAFt 2. ABC ,s nght-angled at B and O is the middle of AC Ar- ^^' ^^ ''''°' °' ""^ •''"''"''e from B. Then 2AC.OD = AB2-BC2. 3. A^C is right.angled at B and, on AC, AD is taken equal ED^=2AE.'dc ' ""^ " ''"'" '^"'^^ ^° ^^- ^^- 4. The square on the sum of the sides of a right-angled tri- angle exceeds the square on the hypothenuse by twice the area of the triangle. 5. To find the side of a square which is equal to the sum of two given squares. 6. To find the side of a square which is equal to the differ- ence of two given squares. 7. The equilateral triangle described upon the hypothenuse of a right-angled triangle is equal to the sum of the equilateral triangles described on the sides BC. Then AC2=2CB.CD. 9. Four times the sum of the squares on the three medians of a triangle is equal to three times the sum of the squares on the sides. I !r ( .: '■ t' I i ' 4'? |j i^n I.. i"^o SYNTHETIC (iEOMKTRV. i.v 14. 10. ABCI) is a rcctanprle and I' is anv point. Then 11. O is the centre of a circle, and AOH is a centre-hnc. ()A-()Ii and C is any point on the circle. Then AC'-'+BC-^ a constant Define a circle as the locus of the point C. 12. AD is a perpendicular upon the line OB. and BE is a ])erpcndicular upon the line OA. Then OA OF = OB.()D. Two equal circles pass each throu^di the centre of the other. If A, B be the centres and E, F be the points of intersection, EF- = 3AB^. If EA produced meets one circle in P and AB pro- duced meets the other in O, PO''*=7AB2. ABC is a triangle having the angle A two-thirds of a right angle. Then AB2 + AC-'=BC-'-f AC. AB. 15. In the triangle ABC, D is the foot of the altitude to AC and E is the middle point of the same side. Then 2ED.AC = AB2-BCa. 16. AD is a line to the base of the triangle ABC, and O is the middle point of AD. If AB-' + BD2=AC2+CD2 then OB -OC. ' 17. ABC is right-angled at B and BD is the altitude to AC Then AB.CD = BD.BCandAD.CB = BA BD 18. ABC is a triangle and OX, OY, OZ perpendiculars from any point O on BC, CA, and AB respectively. Then BX^'+CYHAZ^'=CX2 + AY^'-fBZ^. A similar relation holds for any polygon. 19. AA„ BBi are the diagonals of a rectangle and P any point Then PA-+ PB-'-f PA,H PB,^'=AA,H4P02, where O IS the mtersection of the diagonals. 20. ABC is a triangle, AD, BE, CF its medians, and P any pomt. Then PAHPBHPC-'-=PD2-fPE2+PF2+KADHBEHCF2) 2PA2=ZPDH.\2< where;;/ is a median. ' 21. If O be the ceniroid in 20, AREAL RELATIONS. '31 -5- It A B, C be equidistant points in line md J^ . f u Po-t .„ same line, the d^erence b twecn 1 e J on AI3 and DB is equal to the rect3" ^^e squares CD. rectangle on AD and =6- IfA,Ii, CD be any four points in line AD2+]iO=AC=+IiDH2Au'cD =7. Any rectangle is equal to one-half the ^emnele ,. Middle poi"„'.^tAc'a^dF:;T,r;c/^'^-''= 3. The ..et::;:^ec:.;ira:eir::f ^-- , of the perpendicular from tZ'Llf^ '^ ""= 'r^'" ? be the distance between ZZTr , ''*^°"''' """^ perpendiculars so drl" n, ""' '"° P"""^' M'^+>=«^andj,V,?TK=^2_„2,^ . 30 ABCD'i!'"'"'^'"'"' " ""'"'^^ ^y -'«*+^^ 30. ABCD ,s a square. P is a point in AB produced and O '3 a pomt m AD. If ,he rectangle BP OD '• ~ Stan,, the triangle PQC is constant ' ^ " '""■ If*eIe„gthsof the sides Of a triangle be expressed by K" and Vbe the'de ^''^ 'T''' '' "S^'-gled. ' ..^e altitude":: ^li:^T^^' '^-^'« -'i / Be 31 32 a' A' '% ."N?- t.' 1 ' ■] m !■ t H ri^ ^H i^M j 'fl ,. ■ 132 SYNTHETIC GEOMETRY. 33. The triangle whose sides are 20, 15, and 12 has an obtuse angle. 34. The area of an isosceles triangle is 8^/T^ and the side is twice as long as the base. Find the length of the side of the triangle. 35. What is the length of the side of an equilateral triangle which is equal to the triangle whose sides are 13, 14, and 15.? 36. If AB is divided in C so that AC----2BC2, then ABHBC2=2AH.AC. 37- Applying the principle of continuity state the resulting theorem when B comes to D in (i) the Fig. of 172°, (2) the Fig. of 173°. 38. Applying the principle of continuity state the resulting theorerti when B comes to E in the Fig. of 173°. 39. The bisector of the right angle of a right-angled triangle cuts the hypothenuse at a distance a from the middle point, and the hypothenuse is 2b. Find the lengths of the sides of the triangle. 40. Construct an equilateral triangle having one vertex at a given point and the remaining vertices upon two given parallel lines. 41. A square of cardboard whose side is s stands upright with one edge resting upon a table. If a lower corner be raised vertically through a distance a, through what distance will the corner directly above it be raised } 42. What would be the expression for the area of a rectangle if the area of the equilateral triangle having its side the 71 J. were taken as the 7c,a.} 43- The opposite walls of a house are 12 and 16 feet high and 20 feet apart. The roof is right-angled at the ridge and has the same inclination on each side. Find the lengths of the rafters. 44 Two circles intersect in P and Q. The longest chord through P is perpendicular to PQ. ARKAL RELATIOx\S. ,,- „ T., , ^"^ '"° ^"'^^ Sivcn is isosceles. 48. The largest isosceles triangle with variable base has its s,des perpendicular to one another. " 49. The largest rectangle inscribed in an acute-angled tri- angle and having one side lying on a side of Ih 50 L ^r^''■^^^'^•->''''"^--ha'f•hVo"t;e Lgle ApTi ° ''"t,T''"« i" O.-d P is any pint. Thl , ,."""' '""^ •="«'"£ L in A and M in U The tnangle AOB is least when P bisects AB. EQUALITIES OF RECTANGLES ON SEGMENTS RELATED TO THE CIRCLE. 176°. Theorem, — If two sprantc f« .u sprt f),» » . secants to the same c rcle inter is equal to the coir / ^^''^ '° °"" °^ *^^^ ^^^^nts othe'rsecrnt. '^"'^P^'^^'"^ rectangle with respect to the I. Let the point of intersection be within the circle. Then AP.PB = CP.PD. Proo/.-AOB is an isosceles triangle, and P IS a point on the base AB .-. OA2-OP2=AP.PB. *(i74°N Similarly, COD is an isosceles tri- angle, and P a point in the base CD Bui* 0C2-0P2=CP.PD. C-OA, AP.PB = CP.PD. ^.e.^. I i ! I ! I K^' n [ IK P' 1! ft' 1 , ( ■ m< i . B' ;|: ' Mi f'Tf 1'. 134 SYNTHETIC GEOMETRY. Cor. I. (a) Let CD become a diameter and be ± to AB. (96°, Cor. 5) Then AP . PB becomes AP2, AP2 = CP.PD, and denoting AP by c, CP by v, and the radius of the circle by r, this becomes which is a relation between a chord of a 0, the radius of the 0, and the distance CP, commonly called the verses/ sine, of the arc AB. ib) When the point of intersection P passes without the we have still, by the principle of con- tinuity, AP. PB=CP. PD. But the ns being now both negative we make them both positive by writing PA.PB = PC.PD. Cor. 2. When the secant PAB be- comes the tangent PT (109°), A and B coincide at T, and PA . PB becomes PT2, .-. PT2=PC.PD, i.e., if a tangent and a secant be drawn from the same point to a circle, the square on the tangent is equal to the rectangle on the segments of the secant between the point and the circle. Cor, J. Conversely, if T is on the circle and PT2=PC . PD, PT is a tangent and T is the point of contact. For, if the line PT is not a tangent it must cut the circle in some second point T' (94°). Then PT.PT' = PC.PD = PT2 Therefore PT=Pr, which is not true unless T and T' coin- cide. Hence PT is a tangent and T is the point of contact. Cor. 4. Let one of the secants become a centre-line as PEF. Denote PT by /, PE by h, and the radius of the circle by r. Then PT2=PE . PF becomes t^^h{2r+h). 4. I N AREAL RELATIONS. Exercises. 135 2. 4. mouth. A sphere of radius >• is di-opoed into if h rar . the centre of the sphere .oJ^CZ JZ '' ^'LTventf s"r ''"' """^^^ ^' 7'9^° -"-^ how — •:; ::ntrh;:ir ^^'-^ - ^-- ^^^ to; or . '■ "^T/tet::?";"'/,^" " ^"^ ^ ^-^^heir centres and .1 ^.l f '""* '^' ^'"^^h «^ th^i^ ^«n^mon chord and also that of their common tangent. 9. Two parallel chords of a circle are^- pnd . ^ 1. • distance apart is ^ to find !k /^ """"^ '^'"' circle. *^^ ^^^^"s of the 10. If ., is the versed sine of an arc, l^ the chord of half the arc, and r the radius, /:'^=2vr. 177°- Theo7'em. — If unon pnrh r>r <■ raiV of ,^^- * u V ^ °^ ^^^° intersect ng lines a pair ot pomts be taken mirh ff,^f *u ^ ^ -cnts between the JoLr:' nleclTrd tVe" ""^ ■"^'; PO.nts in one of the hnes is equal tolrco^resptd.^ J, , Mil! '! 'Ill j-'f 1,11 1 1" 1.6 SYNTHETIC GEOMETRY. angle for the other line, the four assumed points are concyclic. -^ (Converse of 176°.) -^ L and M intersect in O, and OA.OB = OC.OD. Then A, B, C, and D are concyclic. Proof.— Since the os are equal, if A and B lie upon the same side of O, C and D must lie upon the same side of O ; and if A and B lie upon opposite sides of O, C and D must lie upon opposite sides of O. Let a pass through A, B, C, and let it cut M in a second point E. Then OA.OB = OC.OE. (176°) ^"t OA.OB = OC.OD. (hyp) OD = OE, and as D and E are upon the same side of O they must co- mcidej .-. A, B, C, D are concyclic. g.e.c^, 178°. Let two circles excluding each other without contact have their centres at A and B, and let C be the point, on their common centre-line, which divides AB so that the difference between the squares on the segments AC and CB is equal to the difference between the squares on the con- terminous radii. Through C draw the line PCD J_ to AB, and from any point P on this ^ ' line draw tangents PT and PT' to the circles. Join AT and BT'. Then, by construction, AC2-BC2 = AT2-BT'2. But, since PC is an altitude in ■BC2=AP2-BP2, AP2=AT2-f-PT2, BP2=BT'2-|-PT'2, P'p2_p'p'2 PT=pr.' (172°, I) (169°, Cor. I) AREAL RELATIONS. n? Therefore PCD is the locus of a point from which equal tangents are drawn to the two circles. nef.~T\ns locus is called the radicai axis of the circles and IS a line of great importance in studying the relations of two or more circles. Cor. I. The radical axis of two circles bisects their com- mon tangents. Cor. 2. When two circles intersect, their radical axis is their common chord. Cor. 3. When two circles touch externally, the common angent at the point of contact bisects the other common tangents. 179°. The following examples give theorems of some im- portance. Ex. I. P is any point without a circle and TT' is the chord of contact (114°, Def.) for the point P. TT' cuts the centre-line PO in Q Then, PTO being a n, (110°) OQ.OP = OT2. (169°) .'. the radius is a geometric mean be- tween the join of any point with the centre and the perpendicular from the centre upon the chord of contact of the point. Def—V and Q are called inverse points with respect to the circle. Ex. 2. Let PQ be a common direct tangent to the circles having O and O' as centres. Let OP and O'Q be radii to the points of contact, and let QR be || to 00'. Denote the radii by r and /. Then AC = 00'4-r-r', BD = 00'-r-f-»-'. .-. AC.BD = 00'2-(r-O2=0R2-PR2=P02 (T69°,Cor. t) 138 SYNTIIKTIC GEOMETRY. Similarly it may be shown that AD . BC = square on the transverse common tangent. ^ EXKRCISES. .. The greater of two chords in a circle is nearer the centre than the other. 2. Of two chords unequp.lly distant from the centre the one nearer the centre is the greater 3. Ali is the diameter of a circle, and P, Q any two points on he curve AP and BQ intersect in C, and AQ and BP in C . Then AI\AC + BQ.BC=:AC'.AQ + BC'.BP. 4. Two chords of a circle, AB and CD, intersect in O and are perpendicular to one another. If R denotes the radius of the circle and E its centre, 8R^ = ABiJ + CD- + 40E'^ 5. Circles are described on the four sides of a quadrangle as diameters. The common chord of any two ad^cent circles is parallel to the common chord of the other two. 6. A circle S and a line L, without one another, are touched by a variable circle Z. The chord of contact of Z passes through that point of S which is farthest distant from L. 7. ABC is an equilateral triangle and P is -ny point on its crcumcircle. Then PA + PB + PC=o, if we consider the line crossing the triangle as being negative. 8. CD IS a chord parallel to the diameter AB, and P is any point in that diameter. Then PC- + PD- = PA-+PB2. CONSTRUCTIVE GEOMETRY. 139 SECTION V. CONSTRUCTIVE GEOMETRY. 180°. Problem.~K^ being p given segment, to construct the segment A 13^2. 6V«j/r. -DrawBC_LtoABandequaltoit. Then AC is the segment AB^/2. Proof.~^\nQ^ ABC is right-angled at B AC2=AB^+BC2=2AB2, (169, cLr. i) AC-ABV2. ^ Cor. The square on the diagonal of a given square is equal to twice the given square. ^ 181°. Problem.— lo construct A Bv/3. C^«j/r.-Take BC in line with AB and equal to It, and on AC construct an equilateral tri- ""?;!^^^5^- (124°, Cor. I) BD is the segment ABV3. Proof.-k^Vi is a H, and AD=AC = '>AB Also AD^ = ABHBD2=4AB^. (,69°, Cor. ,) BD2 = 3AB^,andBD = ABV3. Cor Since BD is the altitude of an equilateral triangle and AB is one-half the side, ^ .-.the square on the altitude of an equilateral triangle is equal to three times the square on the half side. 182°. Problem.— '^Q construct AB^/5. a Constr. -Draw BC ± to AB and equal to twice AB. Then AC is the segment AB^S. Proof.— ^xnQ.^ aB is a right angle, AC-'=AB-^+BC^ But BC-=4AB-; AC-=5AB^ and AC = ABV5. B '! ii I: w -m f 140 SVNTHETIC GKOMETRY. are given. " convenient. A few examples Ex. ,. AB being a g,iven segment, to find a point C in its A B ^"^^ "'•'" AO= AB . CB. Analysis- Aa=An.CB = AB(AB-Aa •• .^ . AC^' + AC.AB=AB=. * '^^^' r ufcT'lr'"^ T '"'" ''^*"'^ '■"^ «"'' -'ving as a quad- 1 .It c m AC, ive have AC = l(AB^c - A Kl "nd Ihis is to be constructed. ^' ^' c' I < I \ C>«/r.-Construct AD=ABV5 (by ,8.-) as in the figure \ \ \ and let E be the middle point of BD \ ' qV TakeDF = DE. ' V ^ Then AF=ABv/5-AB; .-. bisecting AF in G, AG=AC = KAB^/5-AB), and the point C is found. Again, since ^/5 has two signs + or -, and we have AC= -Ua:^^^ ''^ "^^^''^ ^'^° inerefore, for the nnmf r" a t^ l^r=DE, and bisect AplC'Tren '"""''' '^''^ AG'=i(ABV5 + AB); and since AC' is negative we set olT AG' from A to C and C IS a second point. 'i to c, and The points C and C satisfy the conditions, AC==AB.CBandAC'2=AB CB CONSTRUCTIVE GEOMETRY. 141 It is readily proved however. For AD'^=5AB-, and also AD-=(AF + FD)-' = (2AC+AB)2 whence AC^ = AB(AB - AC) = AB. CB. ' It will be noticed that the constructions for finding the two points differ only by some of the segments being taken in different senses. Thus, for C, DE is taken from DA, and for C , added to DA ; and for C, AC is taken in a positive sense el^lt^Al.''^^' '''''^^'''' "'^" - ^ -^-^- -- In connection with the present example we remark :- I. Where the analysis of a problem involves the solution of a quadratic equation, the problem has two solutions corre- spondmg to the roots of the equation. 2 Both of the solutions may be applicable to the wording ot the problem or only one may be. 3- The cause of the inapplicability of one of the solutions IS commonly due to the fact that a mathematical symbol is more general in its significance than the words of a spoken language. * 4. Both solutions may usually be made applicable by some change in the wording of the problem so as to generalize it. The preceding problem may be stated as follows, but whether both sohations apply to it, or only one, will depend upon our definition of the word "part." See Art. 23°. To divide a given segment so that the square upon one of the parts ts equal to the rectangle on the whole segment and the other part, De/.~K segment thus divided is said to be divided into extreme and mean ratio, or in median section. Ex. 2. To describe a square when the sum of its side and diagonal is given. Analysis.-If AB is the side of a square, AB^^-- is its diagonal. ^^^^.^ I'M ... , ' !fl I '42 SYNTHETIC OKOMKTRY. Ali(i + s'2) is a given scRment =.S, say. Then Al) = SfV2-i). r.«,r/,- Let EF be the given segments. I '-aw i a X and = to EF, and with .Tko <•:,'',:;';„"' '^^"'''="®-- square is easny^conslncfed'^ '"' " "" ^''"^^^ • ""-- '"^ diagonal is the .^ n s ' ,enfs The 7 u" ''? ^'"^ ='"" is ve. suggestive, hut w^^f its:^^!^ ll^r o:^i:i::e:^-hX— --^ ^- j'^ ^"^ AC the given rectangle. I On DA produced make AP = S and . . -Q draw PBO to cut DC produced in Q. '-Q-' CQ IS the segment required. ^r../-Complete the os PEQD, PGBA, and BCQF Then c::AC=oGF = GB.BF = PA.CQ, ^ S.CQ=aAC. 185°. Problem.—lo find the siH^ ^f o me side of a square which is F equal to a given rectangle. iu?'"n^~'-" ^'^ "^ "'^ '««a„gle. Make BE = BC and in line with BA. Un AE describe a semicircle, and pro- duce CB to meet it in F. B F is the side of the required square. /''■<'<'/- Since AE is a diameter and FB a half chord X to it, CONSTRUCTIVE GKOMETRY. 143 (176°, Cor. I) BF»=AB.BE, BF2=AB.IiC. Cor. This is identical with the problem, "To find a eeo metru: mean between two ,iven segments," and it tnifhes" - (165°) remngie.'^° '°"''''"'' '" '''""''''■''^^ ^''•^"^'^ ^^"'-^^ *« '-^ g-en i'QK to be the required triangle. Then AB.BC=:^PR.QT ■ ' = PT.QT. But QT=PT,/3, (181", Cor) PT.QT=PTV3 whence PT^ = AB^ 3 . J^BC. - . .. s'ldt T AP';J'' 1^r.T''' '^"'^^ ^° *^^ '■^^^^"^^le whose r27°'and r^S^:^' '""^ ^'''^'^"' ^^ ^"""^ ^^ "^^-"s of 181°, Thence the triangle is readily constructed. itstse.' "^^ ''"''' '"' "'' ''" ^''""^^^ ^y ' ^-^ P--"eI to Let ABC be the triangle, and assume PQ as the required line, and complete the parallelograms AEBC, KFBC, and let BD be the altitude to AC. Because PQislltoAC, BDis±toFO. Now £=7EP = c=7PC, .-. £=7FC = ci:7EO, or PQ. BD=AC. BG. (icr\i d- .^,^^^^^=^EC,or.PQ.BG=AC.BD; ^ '' ' ^ j^^^d.v.dmg one equation ^b^the^other, and reducing to one termined ' ' '"^ '"' ^"^'"^'"^ ^^ ^^ ^^ ^^-- \ iti Mi!' ' 144 SYNTHETIC GEOMETRY. 186°. Problem.-^To find the circle which shall pass through two given points and touch a given line. Let A, B be the given points and L the given line. Constr.-h^i the line AB cut L in O. Take OP = OP', a geometric mean be- *T, u , ^'''^^" ^^ ''^"d Oi^ (iS5°). The circles through the two sets of three points A, B,? Ind A, B, P a e the two solutions. ' The proof is left to the reader. (See 176°, Cor. 2.) I87^ Problenu-To find a to pass through two given points and touch a given 0. Let A, B be the points and S the given 0. G?;w/;-.__Through A and B draw any so as to cut S in two points C and D. Let the line CD meet the Hne AB in O. From O draw tangents OP and OQ to the 0S (114°). Pand the 0s which pass through A^a^d S tuct T Th^^ fore the 0s through the two sets of three pc^n s a' B P "d A, B, Q are the 0s required. ' ' ^^^ Proof,- OB.OA = OC.OD = 002=OP2. therefore the 0s through A, B, P and A, B, Q have OP and to ns 'r'T ^^f' ^°^- '^- ^"^ ^^-^ -^ als'^^ngents "Ssr^-^ ' ^"^ ^ -^ ^^^ p^^- ^^ -tact o? th: 12. Exercises. I. Describe a square that shall have twice the given square. area of a 2. Describe an equilateral triangle equal to a -iven s square. CONSTRUCTIVE GEOMETRY. 4- Construct AB./7, where AH .= o' • 5. Construe. V«^l7 ^,f, '^; ^.ven segment. given line segments ' "'"" " """ ' *"»'« 6. Dmde the segment AU in C so that AC^->CB= 9,, that AC is the dia-oml of ,h. ' ''''°"' .h.hoMrorexterlirauT;"^""^'^- '^°- '^avingAf;e':r:rAOarrad,';:;r„T %\ ''^l "o dil'i de' '™"™"'°" "f "^' -'-^ he problem -. Show that the cons. cZ of ,8 ?";" °T' P'"'^'" 1 1. Construct an equilateral triangle when the sum nf ,-. -^ Descnbe a square in a given acu.e-angled trian-^Ie so >4. W,thin an equilateral triangle ,o inscribe a second eaui lateral ^.r,a„gle whose area shall be one.h:™!:?:^ " '™ra„Vdrr ^/ic^rA^'t ;r r"^ - '"^ given square. ^ '^^" ^^ ^^^^^ *« a K 12. 13. 14^ SYNTHETIC GEOMETRY. ir>. Draw a tangent to a given circle so that the triangle formed by it and two fixed tangents may be (i) a maximum, (2) a minimum. 17. Draw a circle to touch two sides of a given square, and pass through one vertex. (Generalize this problem and show that there are two sohitions. 18. Given any two lines at right angles and a point, to find a circle to touch the lines and pass through the point 19. Describe a circle to pass through a given point and to touch a given line at a given point in the line 20. Draw the oblique lines required to change a given square mto an octagon. If the side of a square is 24, the side of the result- ing octagon is approximately 10; how near is the approximation ? 21. The area of a regular dodecagon is three times that of the square on its circumradius. 22. By squeezing in opposite vertices of a square it is trans- formed into a rhombus of one-half the area of the square What are the lengths of the diagonals of the rhombus? 23. J , Q, R, S are the middle points of the sides AB, BC, CD, and DA of a square. Compare the area o'f the square with that of the square formed by the joins AO BR, CS, and DP. ^ ^ ^' 24. ABCDEFGH is a regular octagon, and AD and GE are produced to meet in K. Compare the area of the tri- angle DKE with that of the octagon. 25. The rectangle on the chord of an arc and the chord of Its supplement is equal to the rectangle on the radius and the chord of twice the supplement. 26. At one vertex of a triangle a tangent is drawn to its cir- cumcircle. Then the square on the altitude from that vertex ,s equal to the rectangle on the perpendiculars from the other vertices to the tangent. 27. SOT is a centre-line and AT a tangent to a circle at the point A. Determine the angle AOT so that AS = AT. PART III. PRELIMINARY. I88°. By superposition we ascertnin n,« equaluy of two given line-segments li T'"^ "' '"' the relation between the lengths "/; " '" '''P''"^ endeavour to find two numefkn, / """""'" ^^^"^"'^ "« another the same relation^ n m^"?','"" "'"^'' •""" "> ""e ments do. '" ""^gn^^de that the given seg- Which the mLines o AB nd^L to''1 "'" "^P-^" '° numbers. Le, ,„ denote the ,r. ^^ ' ? "■' ''''"' ""'"lo me..sure ofCD with respect to ■W-V' ^" "■"• « 'he The numbers m andThlw , """"'^"S"'- tions as to magn ude 'h, , ' '" °"^ -"'h- "-e same re.a- T^ . . „/ '"'^ '^2""^"'= AB and CD do. Ihe fraction _ is caller! ,•„ a ■». of « to „ ,nd -"r '" "'•^'^^''■•^ 'he n,//. "I m to «, and m Geometry it is rili«.i .1 • ■ Now n has to m the same ratio ^ f ° "^^^ "' ^D. -• But if CD be ,1 """^ ^"^ '° 'he fraction „ CD be taken as ../. its measure becomes unity while that ofAB becomes 2-'. Therefore the w/a of AB"to CD i. tJ,» respect to CD as unit-length "'"""•^ °^ ^B with ill 148 SYNTHETIC GEOMETRY. oni'vtf " ""k r*^T"" "' '"^"""ensurable the ratio can only be symbohzed, and cannot be expressed arithmetically except approximately. ^ unlilT; J^""' '"P^°'^ ^^ '° '^^ '"P"^^^ °^ '^^•"^ stretched until It becomes equal in length to AB, the numerical factor which expresses or denotes the .mount of stretching neces- sary may conveniently be called the tensor of AB with respect to CD. /rr , s Ac. f X (Hamilton.) As far as two segments are concerned, the tensor, as a numerical quantity, is identical with the ratio of the segments bi^ It introduces a different idea. Hence in the cLeTcom.' rutTth. "'T'' '^" ''"^"' " arithmetically expressible, but m the case of incommensurable ones the tensor maybe symbolically denoted, but cannot be numerically expressed except approximately. p*c:,bcu Thus if AB is the diagonal of a square of which CD is the side, AB = CDV2 (180°); and the tensor of AB on CD ...., the measure of AB with CD as unit-length, is thai numerical quantity which is symbolized by ^2, and which can be expressed to any required degree of approximation by that amhmetical process known as "extracting the square 190°. That the tensor symbolized by ^2 cannot be ex- pressed arithmetically is readily shown as follows •- If x/2 can^be expressed numerically it can be expressed as a fraction, ^, which is in its lowest terms, and where accord- ingly m and « are not both even. If possible then let sj2 = ^. Then 2ir=m\ Therefore vr and m are both even and ;/ IS odd. But if m is even, ^- is even, and ifi and ;/ are both even. But n cannot be both odd and even. - ..eretore y 2 cannot be arithmetically expressed. PRELIMINARY, 149 //W^//.« of an incommensurable tensor. > . E'F' Then some tensor will bring AB to AC Let BD be divided into 10 equal parts whereof F nn.1 r arethosenumbe:^d4and5 ts wnereot E and F .he second too ,.a.. an^^ ttjee:^::^^ ^"^" ^'^ a.fZ;n^,r :^,^ r '° -- -- --or K; . too grea? ' "' ''" "^^'"^ "° -"''" -"<• 'he second - Srrr^':^^^^^^^^^^^^ - oM.n sides of C and adjacent to ft. P°" °PP°="^ Thus, however far this process be carried, C will alwav, i;» ^./w.« two adjacent ones of the points toobtTilled '^'^ '" the nlTthTf ""°" '"" '"'"^P^"^- °-'»'h of tne length of the former ones, we may obtain a noint „f division lying as near C as we please ^ Now if AB be increased in length from AB to AD it must at some period of its increase be equal to AC Therefore the tensor which brings AB to AC i, , r.=i The preceding illustrates the difference between magnitude and number. The segment AB in changing to AD passes r nl'eS '"'""''1? '^"^"'- ^"' '^e'commensu aS mustproceedh ■"■''""' .^™"""'' 'y'"= ''«"«" ■ -"d ^ iiiusc proceed bv snmp unif }T^"'-, ,- n j not continuous. ''"^"' ^'"^ "'^ '^^"'^'"^ in . I n |'« ; ^1' ISO SYNTHETIC GEOMETRY. Hence a tnagnittide is a variable which i,, ^n^ • r onevalu. . ano,.r^ passes /W^. jj t^^Xf^r AC to AB numerical factor which brings But according to the operative principles of Algebra AB ^ ' AB. AC .AC=AB, ••• AC is the tensor which brings AC to AB ratio of the numerator to the denominator. SECTION I. PROPORTION AMONGST LINE-SEGMENTS. l')^\ Def.-Four line-segments taken t„ order form „ proportion, or are in proportion, when the tensTrofZ^ onthe^eond is the same as the tensor ^Zt^^lfX This definition gives the relation a __ c where «, l,,c, and ^denote the segments taken in order The fractions expressing the proportion are subjecf o all the transformafons of algebraic fractions (,58') and the re u .s geometrically true whenever it admit o rgeomet" interpretation. geometric The statement of the proportion is also written ""''^-'-'^^ (B) I'KOPORTION AMONGST UNE-SEGMKNTS. .5, Tn'l'Tf \"*'" •■ '■"'"'^'"" «"= '^'"^ion of the quantity de In either form the proportion is read "^ is to (^ as ^ is to ^t'." 193°. In the form (b) a and ./ are called the .r/remcs ind a\ c ;«:''" ^""'^"^^^ ' ^" ^^- '^^ °PPosites of the 194^ I. From form (a) we obtan. by cross-multiplication wJiich states geometrically that ;^/..« four segments are in proportion the rectanol^ and this equality can be expressed under any one of the fol lowmg forms, or may be derived from any one of them v.^' «^t «=l' !'_b' b a! . a! b' b' b' a'~2' b'~-i' .n all of vvh,ch the opposites remain the same. Therefore 3 Two equal rectangles have their sides in propo,t!on l\ III l[ in I-', ri?« I w i:fi j r 1 "' 1' 1 J s '1 !lli .11 J! 52 SYNTHETIC GEOMETRY. 195°. The following transformations are important. Let "='. then b d I. 2. q±^^c±d b d ' a_c_a+c a-c {a>b for - sign) b'd-TTd^J^' (^>^for - sign) ^^' rrr^'''- '^^'^ ^ ^ f _ ^ _ « +/+J? + etc. b d y <^ + ^/+7feTc.' To prove i. •.• a 1)' d b d ^ and a±b b c±d d ' To prove 2. *.• a b c . a b d' ' ' Td' (194°, ?. a±c b±d c d ' or a_c a±c b d b±d' To prove 3. •.• a T c a + c P ^ f f+Q, f+d+J ., etc. SIMILAR TRIANGLES. 196°. Dcf.—i. Two triangles are swnVar when the angles of the one are respectively equal to the angles of the other. (yf, 4) 2. The sides opposite equal angles in the two triangles are corresponding or homologous sides. The symbol « will be employed to denote similarity, and will be read "is similar to." A' D' C PROPORTION AMONGST LINK-SEGMENTS. 1 53 In the triangles ABC and A'B'C, if z.A=i.A' and lB= ' B' then also lC=lC and the tri- angles are similar. The sides AB and A'B' are homologous, so also are the other pairs of sides opposite equal angles. a d Let BD through B and B'D' through B' make the ^BDA=^B'D'A'. Then AABD « AA'B'D' since their angles are respectively equal. In like manner ADBC « AD'B'C, and BD and B'D' divide the triangles similarly. 3. Lines which divide similar triangles similarly are homologous lines of the triangles, and the intersections of homologous /mes are homologous points. Cor. Evidently the perpendiculars upon homologous sides of similar triangles are homologous lines. So also are the medians to homologous sides ; so also the bisectors of equal angles in similar triangles ; etc. 197°. Ty^^^rm.— The homologous sides of similar triangles are proportional. AABC^AA'B'C having lA = lA' and z.B=^B'. A^ = B^C^ CA A'B' B'C C'A'* Then B c Proo/.-Vlace A' on A, and let C fall at D. Then since LA =z.A, A'B' will lie along AC and B' will fall at some point 1^. Now, AA'B'C'^AAED, and therefore z.AED=z.B and B, D, E, C are concyclic. /j^^ov' Hence AD.AB = AE.AC, (176° V^ ^^ A'C'.AB=A'B'.AC. AB AC 11 r ill Ml ll F V 1 t 1 1 ^ 1 i , ! 1 if ' . ! i 1 'C il iWl A'B' A'C* (194°, 2) '54 SYNTHETIC GEOMETRY. Similarly, by placing B' at B, we prove that AB^ BC A'B' Wc' AB _ BC ^ CA A'B' B'C CA'" q.c.d. Cor. I. Denoting the sides of ABC by a, b, c, and those of a' V~c'' Cor. 2. i^ = ^ = ^_ ii-^b-\-c . „ a! b' 'c~a' + b'-^c" (^95,3) t.c., the perimeters of similar triangles are proportional to any pair of homologous sides. 198°. Theorem.—Tv^o triangles which have their sides pro- , B' portional are similar, and have B A their equal angles opposite hom- / \ ologous sides. (Converse of 197°.) Af )c' AB^BC^^CA A'B' B'C C'A'' Then ^=zA', ^B=z.B', and Proof.-On A'C let the AA'DC be constructed so as to have the and Then and but m • • and and • • and z.DA'C'=.lA ^DC'A'-^C. AA'DC ^AABC, AB^AC^BC A'D A'C DC' AB ^ AC ^ BC A'B' A'C B'C ' A'D = A'B' DC = B'C, AA'DC = AA'B'C'. lM=lA, z.B'=^B, (196°, DtL I) (197°) (hyp.) (58°) q.e.d. PROPORTION AMONGST LINE-SEGMENTS. 1 55 i99\ Theorem.~.\i two triangles have two sides in ^.rh proportional and the included ^^^^ angles equal, the triangles are similar. AB_AC ^ , A'iy~"A'C" ^'^^ '^=^, then AABC^AA'B'C. A'rT^T^^'?/' "" '^' ''^"^^ ^'^ A'C' li^ along AB, and A B he along AC, so that C falls at D and B' at E The triangles AED and A'B'C are congruent and therefore similar, and AB^AC AE AD Hence AB.AD^AE.AC; /,q.°^ and ... B, D, E, C are concyclic. \^1}. z.AED = ^B,and^ADE=^C, (106° Cor , and AABC^ AAED^AA'B'C. ' ,3. 20o\ Theorem.~\{ two triangles have two sides in earh LTe^^al:' ^"' ^" ^""^'^ '''-'' ^ homologt 'sidT in trilnj^f :r::t^^^^^^^ ''' ^^"^^^ °^ ^^^ -° ^^^- the tH^nXf :::'t^^r ;:: ^^-^- ^^ ^^^ - ^^^- the similar. AB_BC , ^ ^ A'B'~B^C'^^^^=^'- dA \ ^' I- If BOAB, AABC^AA'B'C. /^;-../_Place A' at A and let B' fall at D, and A'C along AC. Draw DE II to BC. Then AABC^ AADE, and ^^-= ^^ But AB^BC AD B'C ' AD dp:' DE = B'C'. M H f 1 *! ft 156 syntiib:tic geometry. And since B'C> A'li', the AA'B'C'^ AADE and they are therefore similar. /^^o x •n (05 , I) But AABC^AADE, AABC^AA'B'C. 2. If RC- the figures! " '"'''' ""' """'"' "' ^^"-"'«^'-- lines of 207°. 7y/.w;;/.--.Two similar rectilinear figures have any two hne-segn.ents from the one proportional to the homolo- gous segments from the other. Proof. ~^y definition AP^AP', and they are similarlv placed, .-. AE : A'r:' = AC : A'C. For like reasons, AD : A'D' = AC : A'C'*=AB : A'lr AE^AD_AC__AR A'E' A'D'~A'C""A'ir"^^'^-' and the same can be shown for any other sets of homolo.^ous Ime-segments. ^ fi ures '' '^" '''^'"^''' polygons of the same species are similar rUOPOKTION AMOKOST r,,N,,..sKc;Ml.:NT.S. ,6, Now, let (7 fi c i' /' J Similar rcLnilar no'lvrroL* V,""' ''^"""'"Poiis sides of two r a' // / •••-,V + // + ,,^ ; (.95", 3) _ perimeter of I» their radii. ^ ^^lXq^ are proportional to C-or. 3. If ,-, .' ,eno.e .he circun.fcenccs „f two d,Ces and '■■'.mir' .her radii, ;.;, = eons.a,u. Oenote this constant by 2r, then .he appro^in,:.:";;;;; r ;;;:;— t,-^^ '--''>•. (f?-^r^r,:s:hrci:[:s:-:^:- whose radius is k the ton«„- ■' • ,. '^''^ • '«"«» ^ var,es directly as . varies, an also var.os directly as the angle at the centre varies Hence ^ ,s taken as the ,.u-as,.r. of the a„«,e, subtended by the arc, at the centre. Denote this angle by .. Then and when s=r, becomes the ',ni, an.de. inl'^'«.-The bisectors of the vertical angle of a tnangle each divides the base into parts which are propor- B tional to the conterminous sides. BD and BD' are bi- sectors of ^B. Then AD^AD'^AB I^C CD' BC' ^roo/.-Through C draw FRF' -I. o, , EE'lltoAB. Then EBE =-] (45 ), and ^E=^BD = ^DBC BC = EC = CE' But ABD and ABD' are triangles having EE' /f^V^ common base AB ^ ^ " *° ^^^ AB^AD . AB AD' EC DC'-^^d^E'-CD" or AD^AD' AB DC CD'"'BC Cor. D and D' divide the base internally and externally in (203% Cor.) g.e.d. PROPORTION AMONGST LINE-SEGMENTS. 163 209".. Theorem.~-^^ line through the vertex of a trianele d.v,d,ng the base into parts which are proportional. o he contermmous sides is a bisector of the vertical angle (Converse of 208 .) ^^wi-ai diigie. Let the line through B cut AC internally in F. Then, AD being the internal bisector |^=AD ^^^3.^^ ^^^ aB^AF ^^ hypothesis, .*. AF^AD FC DC' But AD is < AF while DC is > FC. tl.:'"r'^^- '^If'u" '' ^"^P««^ible unless F and D coincide ie the line is the bisector AD. "'"^lue, i.e., ext^e^allvlt i'sTh 'b' "'""f A'^' " ""^ ''™ "'"''^^ '"e base externally it is the bisector AD'. 2io^ rW.;«.-The tangent at any point on a circle and he perpendicular from that point upon' the diameter divide the diameter harmonically. ^ ^ ^ AB is divided harmonically in M and T. ^ /'r^^/-z.CPT=^PMT = -l, (iro°) ACPM^^CTP, CM Cp' and = _ ^^ CM CB CP cr ^'^CB^^CT' CB + CM_CT + CB AM AT CB-CM CT-CB' "'^ MB^BT' .'. AB is divided harmonically in M and T. 21 r. The following examples give important results. ('95°, 2) q.e.d. ! ' ; ',' ■II f s ' ( 164 SYNTHETIC GEOMETRY. Ex. r. L, M, and N are tangents which touch the circle at similar, .-. But TA^TP_TB A, B, and P. AX and BY are J.s on N, PC is i. on AB, and PQ and PR are ±s upon L and M. Let N meet the chord of contact of L and M in T. Then the triangles TAX, TPC, TBY are all TA^TB^TP2 AX. BV~Pca" (176°, Cor. 2) (A) (114°, Cor. I) AX PC "by' ■ TA.TB = TP2, AX.BY = PC' Again, let L and N intersect in V. Then VP = VA, ^VQP=z.VXA = n, and ^QVP = ^XVA. AVXA = AVOP, 'ind AX = Pp. Similarly BY = PR, .-. PQ.PR = Pc-. („) Kx. 2. AD is a centre-line and DQ a perpendicular 'to it, ''^"^ AO is any line from A to the line DO, Let AG cut the circle in P Then AADQ^^^^aPB, AD^AO AP AI? or AD.AB = AP.AO. -D But *he circle and the poiln D being given, AB. AD is a given constant. _, • • AP. AO = a constnnf Conversely, if Q moves so that the oaP. AO rem-xi „ p. PROPORTION AMONGST LINE-SEGMENTS. ,65 Now, let the dotted lines reorespnt v.v.vi j , metal jointed toirethei- ,n 17 f ^ '■°'^' "^ """"^ <"■ the PC nts X CPU V .nH n / '''^ ™''''"°" =''^™' rhombus (8 ", Def 'a^d Pt't'Iv"'" "'^' ^'^^^ '^ - being fi J. ' '' '^' ""'' ^"=AV, and AC = CP, AC PQ is the right bisector of UV nnH A Jc « -j- U and V. Therefore A P n / equidistant from Also PTin ' ' ^ ^'^ •'^^^^y^ ^" l'"e. UP being constants, AP Valf ^^n^J^ ' ""' "^ •^"'' And AC being fixed, and CP beinc ecmnl tn Ar i, on the circle through A having C as ce^e ""' ^ "°^" .'. Q describes a line ± to AC bap\4l°s=^;i--- -7J,: o^ -ch each bist«fhf:A'*'=''™°"'^'^<'"'-'''-->'"'°'et'AD Then z_B=^BAD = / DAT o«^ r- - ;o .he ..angles ABC airrfic^Tlfer^ X: Also, AABD is isosceles and AD = DB=AC BA:AC=.AC:DC, "•* I3C:BD = BD:DC •• , ^ BC.DC = BD2 And BC IS divided into extreme and mean ratio at D (r8.° Ex. I). Thence the construction is readily obtained ^ ' Cor. I. The isosceles trian^de APR i,oo i angles eqna. to one-third its v^rtic .u:gt '"' °' "' '"^' with their vertices at 'III If a m form a regular decagon. (132°) 1 66 SYNTHETIC GEOMETRY. Wlli 1' AABC meet its circumcircle in two points ivhich, with the three vertices of the triangle, form the vertices of a regular pentagon. (3) The ^BDA = the internal angle of a regular pentagon. 2 1 If. The following Mathematical Instruments are im- i P portant :— I. Proportional Compasses. This is an instrument primarily for the purpose of mcreasing or diminishing given line-segments in a given ratio ; U, of multiplying given line- segments by a given tensor. If AO = BO and QO = PO, the triangles AGE, i'Oq are isosceles and similar, and AB:PQ==OA:OP. B Hence, if the lines are one or both capable of rotation about O, the distance AB may be made to vary at pleasure, and PQ will remain in a constant ratio to AB The instrument usually consists of two brass bars with slots, exactly alike, and having the point of motion O so arranged as to be capable of being set at any part of the slot. The poinis A, B, P, and Q are of steel. 2. The Sector. This is another instrument which pri- marily serves the purpose of increasing or diminishing given line-segments in given ratios. This instrument consists of two rules equal in length and jointed at O so as to be opened and shut like a pair of com- passes. Upon each rule various lines are drawn corresponding in pairs, one on each rule Consider the pair OA and OB, called the "line of lines " H-ach of the .. of this pair is divided into lo equal parts im- PROPORTION AMONGST LINE-SEGMENTS. 167 which are again subdivided. Let the divisions be numbered numbered 6 are the pomts P and Q. Then OAB and OPO are simihr triangles, and therefore PQ:AB=OP:OA. But A 7^ u •'• PQ = i"oAB. And as by opening the instrument AB may be made equal to can find PQ equal to /^ of any such given segment. lAB VT'' Tr^"^''"'^"" ^^" '^°^ *^^^ t^^ distance 5-5 is 7 7, etc Hence the instrument serves to divide any ^i^en bef fs's ch asb"! ""'":' ^^"^^ P^^^^' ^^^^^^^ ^^^ -- t)er IS such as belongs to the instrument. The various other lines of the sector serve other but verx- smiilar purposes. ^^ 3- 77/^ Pantagraph or Eidograph. Like the two preceding in- struments the pantagraph pri- ^ marily increases or diminishes ' segments in a given ratio, but unlike the others it is so ar- ranged as to be continuous in its operations, requiring only one setting and no auxiliary instruments. It is made of a variety of forms, but the one represented in the figure is one of the most convenient bafs AE tfr Z'' '" ''r ''" J'""^^' ^^ ^ -d B. The bar AE and BF are attached to the wheels A and B respec- tively, which are exactly of the same diameter, and around which goes a very thin and flexible steel band C parallel thryremam parallel however thev be situated w.>i. respect to AB. E, F are two points adjustable on the bars' '■ I. :!| 'I I 'f ft' • f' In 168 SYNTHETIC GEOMETRY. Evidently the triangles DAF ind nnir similar however the incf '^'^^\'^"^ ^^^ remain always 'th fi ""^ "'' "■" '° " '"^ -"- const„^f;tio'™ ■■ '' -suits offer so„,e interesting geo,;etricaU::;::: '" "™ "= 4 T ? j — " A 4. The Diagonal Scale. ^ ==tl±IZ' ™^ ^s ^ flivided scale in \^ which, by means of similar tri- angles, the difficulty of readin<. ,^^^^^_l__n; f n^'nute divisions is very much K^'ZII1'ZIZ.'IZZ_~^\^ dinunished. ^^^^~^~^^B ,^ ^'l '""P^^^ ^'^^"^ is illustrated in A ^^^ figure. A scale divided to fortieths of nn inr-i, ; PROPORTION AMONGST LINE-SEGMENTS. 169 Xtch,"" '' ''^'""" °°' ^"<^ 0«. - AO'B, tha. is on1:tK.c"^"^''' ™ ''■^ ''— ^' ''"- '^ « '-,,, Hence from^ .0 g is one inch and seven-fortieths. In a sun.lar nianner diagonal scales can be n,ade to divide ^ny assumed una-lengd, into any required number of nl^t The chief advantages of such scales ar^ tT^.f ,\ I Exercises. I. ABCD is a square and P is taken in BC so that PC is one.th,r > 'fliP'so„e;nhofBC,whatpartofDBisOI.-P 4. G.ven three line-segments to find a fourth, so' that the four may be in proportion '■ '"'connTf °" u" "''•■'"'^^ °'^ P°'"' '-"d its chord of contact from the centre of a circle is equal to the square on the radius of the circle 6. OD and DO are fixed lines at right angles ,nnd O is a fi'^ed po,,n. A fixed circle with centre on OD and passing through O cuts Ou in P. Then 01' GO k a constant however OCj be drawn -^Z'sj. 'Vans^' a given segment into a given number of equal T,vo secants through A cut a circle in B, D, and C E mihr'^t /"'" ": '"""^'^^ ^""^ -^ ^^" -^ smilar. So also are the triangles ABC and AF D I". Two chords are drawn in a circle. To find a point on 7- 8. 9- I ' 1I 'i I I 3 > f 170 SYNTHETIC GKOMETRV. »«K|,. II. 12. 13. 14. BO is a median. " " ""'i -^^ .ntersect in O. Then If BO, in II, cuts DE in P and AC in n nr, ■ _,. harmonically by P and Q ^' ° " '^""^^'^ In the triangle ABC, BD bisects thf^ / R , D. Then BD^AB. BC Id DC ".p"'? '''' '^ circumcircle.) ^i^.uc. (Employ the 15- ABC is right-angled at B anH tm^ • *i. , . (') The As ADB and 'iUC " '! ""k' '""'"""^ °" ^C. fo\ Qu^ u ^ ^^^ ^''^ch smi lar to AFU' (2) ^how by proportion that AB2=.- AD .^ ^°^^^- and Tjiv> * ^ ' 16. If R and . denote the radii of th^" ' '^^■ Circe of a triangle.^Rt :'.+ ):r"^"* '"' '"" In an equdateral triangle the square on the side is . , to s,x tm,es the rectangle on the radii of tt ^ ' circle and incircle. ""*= '="C"m- °t°h;°the":: "T ''""' ^ ""^ "■«■•"? -"em may be bfsectld b^OB "'"^'' '^^'"•^^" °^ -'^ ^C What is the measure of an angle in rad;,n= , • measure in degrees is 68' 17' ? "' '*^" "^ Jo"^^tude but differing i6MnlLude^'""" 33. ABCDE is'a re^gl;;!:!"' °' ^° ^^'^^' ^^^^ ^^^-. (0 Kvery diagonal is divided into extreme and mean ratio by another diagonal. 17. 18. 19. 20. 21. =5- 2(>. 27- PROPORTION AMONGST I.INE-SKCMKNTS. ,7, (2) The diagonals enclose a second reRular pentaeon Convpare .he s,de and the areas of the two pemagonTof "rnn1t,°'"''"\^'-''"'«'"' '"••■"e'^ ''^ » "^^■•>" proper, t^nal betu-een the other side and the hypothenL, .he al ttude from the right angle divides the hypo.henuse into extreme and mean ratio. omenuse "" Inttnd'x 'TT, " '"'" P""' ^ "«'= '■• fi-d circle '^ + c^ - 2dc cos A. D' A D~c When B comes to B' the ^A becomes obtuse, and cos A changes sign. (214° 2) If we consider the cosine with respect to its magnitude only, we must write + before the term 2^f cos A, when A be- comes obtuse. But, if we leave the sign of the function to be accounted for by the character of the angle, the form given IS universal. Cor. I. ABCD is a parallelogram. Consider the AABD, then BTi- = a' + l?'-2adcosd. Next, consider the AABC. Since Z.ABC is the supplement of d, and BC = AD=(5, AC^=d^ + b'^^2abcosd. and writing these as one expression, (i'^+b^±2adcos9f gives both the diagonals of any CZJ , one of whose angles is 6, Cor. 2. I)E = ,^cos^ (CE being J. to AD), CE=^sin^. AE = ^ + <^?cos 6 ; a sin 6 and tanCAE = ^? = AE d + aco^e' which gives the direction of the diagonal. 218°. Def.-The ratio of any area X to another area Y is the measure of X when Y is taken as the unit-area, and is accordingly expressed as ^. (Compare i88°.) I. Let X and Y be two similar rectangles. Then X=a/f FUNCTIONS OF ANGLES.— ARE AL RELATIONS. 175 and \=a'b\ where a and b are adjacent sides of the aX and a and b' those of the cnY. X_/? b \~a''b'' But because the rectangles are similar '^ = ^ Y rt'^' />., the areas of similar rectangles are proportional to the areas of the squares upon homologous sides. 2. Let X and Y be two similar triangles. Then X = ^r7^sinC, Y=W<^'sinC, Y a'b' Z^' because the triangles are similar, (197°) i.e., the areas of similar triangles are proportional to the areas of the squares upon homologous sides. 3. Let X denote the area of the pentagon ABCDE, and Y that of the similar pentagon A'B'C'D'E'. Then A^ r\ ,' B' P_AD2 R AC2 F" "A'D'2' R' A'C^' /p \ Q y Q_ DC^ ^\\/ Q' D'C'2' ^"-A/ But and D (X) P_R_Q_P+Q+R X F R' O' F + Q' + R'~Y' AD _ AC DC A'D' A'C' D'C" • X DC2 Y D'C'^" (I95^ 3) (207°) And the same relation may be proved for any two similar rectilinear figures whatever. .*. the areas of any two similar rectilinear figures are pro- portional to the areas of squares upon any two homologous lines. m 1/6 SYNTHETIC GEOMETRY. 4. Since two circles are always similar, and are the limits of two similar regular polygons, .-. the areas of any two circles are proportional to the areas of squares on any homologous chords of the circles, or on Ime-segments equal to any two similar arcs. 5. When a figure varies its magnitude and retains its form, any smiilar figure may be considered as one stage in its variation. Hence the above relations, i, 2, 3, 4, may be stated as follows : — The area of any figure with constant form varies as the square upon any one of its line-segments. Exercises. 1. Two triangles having one angle in each equal have their areas proportional to the rectangles on the sides containing the equal angles, 2. Two equal triangles, which have an angle in each equal, have the sides about this angle reciprocally propor- tional, i.e., a : a'=b' : b. 3. The circle described on the hypothenuse of a right-angled triangle is equal to the sum of the circles described on the sides as diameters. 4- If semicircles be described outwards upon the sides of a right-angled triangle and a semicircle be described in- wards on the hypothenuse, two crescents are formed whose sum is the area of the triangle. J. AB is bisected in C, D is any point in /^^^^^^Z^\ AB, and the curves are semicircles, f q \^A/^~^ Prove that P + S = O + R. i cdb ). If a, b denote adjacent sides of a parallelogram and also of a rectangle, the ratio of the area of the parallelogram to that of the rectangle is the sine of the angle of the parallelogram. 14. 15 FUNCTIONS OF ANGLES.-AREAL RELATIONS, i;; 7. The sides of a concyclic quadrangle are a, b, ,, d. Then the cosine of the angle between a and b is 8. In the quadrangle of 7, if ^ denotes one-half the perimeter ^r^^ = ^{{s~a){s~b){s-c){s-~d)], 9. in any parallelogram che ratio of the rectangle on the sum and differences of adjacent sides to the rectangle on the diagonals is the cosine of the angle between the diagonals. 10. Ma, bhc the adjacent sides of a parallelogram and e the angle between them, one diagonal is double the other when cose = JJ- + ^\ io\b a)' If one diagonal of a parallelogram is expressed b> V V «'+ r 7 ' other diagonal is n times as long. Construct an isosceles triangle in which the altitude is a mean proportional between the side and the base Three circles touch two lines and the middle circle touches each of the others. Prove that the radius of the middle circle is a mean proportional between He radii of the others. In an equilateral triangle describe three circles which shall touch one another and each of which shall touch a side of the triangle. In an equilateral triangle a circle is described to touch the mcircle and two sides of the triangle. Show that Its radius is one-third that of the incircle. II 13 41 M PART IV. SECTION I. GEOMETRIC EXTENSIONS. 220°. Let two lines L and M passing through the fixed points A and B meet at P. When P moves in the direction of the arrow, L and M approach towards parallehsm, and the angle APB diminishes. Since the ^ ^ '^ ^"~'~~"^-- lines are unlimited (21°, 3) P may re- cede from A along L until the segment AP becomes greater than any conceivable length, and the angle APB becomes less than any conceivable angle. And as this process may be supposed to go on endlessly, P is said to "go to infinity" or to "be at infinity," and the ^APB is said to vanish. But lines which make no angle with one another are parallel, .*. Parallel lines meet at infinity^ and lines which meet at infinity are parallel. The symbol for " infinity "' is 00 . The phrases "to go to infinity," "to be at infinity," must not be misunderstood. Infinity is not a place but a property. Lines which meet at 00 are lines so situated that, having the same direction they cannot meet at any finite point, and therefore cannot meet at all, within our apprehension, since every point that can be conceived of is finite. 178 c o t( a ir re C( (;eometric extensions. 1/9 The convenience of the expressions will appear throughout the sequel. Cor. Any two lines in the same plane meet : at a finite point if the lines are not parallel, at infinitv if the lines are parallel. 22 1 ^ L and M are lines intersecting in O, and P is any point from which PB and PA are || respectively to L and M. A third and variable line N turns about P in the direction of the arrow. I. AX. BY = a constant (184') = U say. When N comes to parallelism with L, AX becomes infinite and BY becomes zero. ••- ^.o is indefinite since U may have any value we please. 2. The motion continuing, let N come into the position N'. Then AX' is opposite in sense to AX, and BY' to BY. But AX increased to 00, changed sign and then decreased ab- solutely, until it reached its present value AX', while BY decreased to zero and then changed sign. .-. a magnitude changes sign when it passes through zero or infinity. 3. It is readily seen that, as the rotation continues, BY' in- creases negatively and AX' decreases, as represented in one of the stages of change at X" and Y". After this Y" goe- off to CO as X" comes to A. Both magiiitudes then change si-n again, this time BY" by passing through co and AX'^ by pass- ing through zero. Since both segments change sign together the product or rectangle remains always positive and always equal to the constant area U. r- 3i i8o SYNTHETIC GKOMKTRV. 222°. A line in the plane admits of one kind of varia- tion, rotation. When it rotates about a fixed finite point it describes anodes about that point. Hut since all the lines of a system of parallels meet at the same point at infinity, rota- tion about that point is equivalent to translation, without rotation, in a direction orthogonal to that of the line. Hence any line can be brought into coincidence with any other line in its plane by rotation about the point of intersection. 223". If a line rotates about a finite point while the point simultaneously moves along the line, the point traces a curve to which the line is at all times a tangent. The line is then said to envelope the curve, and the curve is called the en- velope of the line. The algebraic equation which gives the relation between the rate of rotation of the line about the point and the rate of translation of the point along the line is the intrinsic equation to the curve. 224°. A line-segment in the plane admits of two kinds of variation, viz., variation in length, and rotation. If one end-point be fixed the other describes some locus depending for its character upon the nature of the variations. The algebraic equation which gives the relation between the rate of rotation and the rate of increase in length of the segment, or radius vector, is the /^/^r equation of the locus. When the segment is invariable in length the locus is a circle. 225°. A line which, by rotation, describes an angle may rotate in the direction of the hands of a clock or in the con- trary direction. If we call an angle described by one rotation positive we must call that described by the other negative. Unless con- venience requires otherwise, the direction of rotation of the hands of a clock is taken as negative. GKOMKTKIC EXTENSIONS. i8i An angle is thus counted from zero to a circumangle either positively or negatively. The angle between AB and A'lV is the ^^' rotation which brings AB to A'B", and IS either +a or -^, and the sum of these A two angles irrespective of sign is a cir- cumangle. /b' When an angle exceeds a circumangle the excess is taken m Geometry as the angle. Ex. QA and OB bisect the angles CAB and ABP extern- ally ; to prove that z_P = 2^Q. The rotation which brings CP to AB is -2a, AB to BP is +2p, :. Z.P = 2(i3-o). Also, the rotation which brings ^ AQ to AB is -o, and AB to BQ P is +p, :. z.Q=^-a. ^(CP.BP) = 2^(AQ.BO). This property is employed in the working of the sextant. ^ 226°. Let AB and CD be two diameters at right angles 1 he rectangular sections of the plane taken in order of positive rotation and starting from A are called respectively the first, second, third, and fourth quadrants, the first being AOC, the p second COB, etc. The radius vector starting from co- incidence with OA may describe the positive ^OP, or the negative ^AOP'. Let these angles be equal in absolute value, so that the AMOPs AMOP', PM being _L on OA. Then PM=-P'M, since in passing from P to P', PM passes through zero. sinAOF = .P;M=-™=_sinAOP. r ft 1 82 SVNTIIKTIC (;i<:OMKTRY. and 01*' OP ^"^^v^i • /. the sine of an an-le changes sign when the angle does, but the cosine does not. 227" As the angle AOP increases, OP passes through the several quadrants in succession. VVhen OP lies in the rst a, sin AOP and cos AOP are both positive ; when OP lies in the 2nd Q,, sin AOP is posi- tive and cos AOP is negative ; when OP lies in the 3rd O the sine and cosine are both negative ; and, lastly, when Op' hcs m the 4th (7., sin AOP is negative and the cosine positive. Again, when P is at A, ^AOP-.q, and PM=o, while OM-^OP. .•• sino=noand coso=i. When P'comes to C, PM = OP and OM==o, and denoting a right angle by J, ^.^^o^ ^^^ ^^ sin J = I, and cos = 0. VVhen P conies to B, PM =0 and OM --- - i , sin 7r = o, and cos 7r= - r. Finally when P comes to D, PM= -OP and OM=o. sin^^'^=:-i, and cos^'' = o. 2 These variations of the sine and cosine for the several quadrants are collected in the following table :— Sine, . Cosine, Sine, . Cosine, ISt (?. + + 2nd Q. + rroin To From To I I 1 o o O 1 I ' 3rd Q. 4th Q. — — + Fiiiiii To Ironi To - I - I - r I 1 GEOM K/rkIC KXTKNSIONS. i«3 228'. ABC is a triangle in its circumcircle whose diameter we will denote bv it ^ ^ Let CD be a diameter. Then z.U = _A, (Io6^ Cor. i) and ^CBD=~1- Al CH = CDsinCI)H=^sinA = ^. and from symmetry, /7 ^_/; sin A sin B c sin C' Hence the sides of a triangle are proportional to the sines of the opposite angles ; and the diameter of the circumcircle is the quotient arising from dividing any side by the sine of the angle opposite that side. PRINCIPLE OF ORTHOGONAL PROJFXTION. 229°. The orthogonal projection (167", 2) of PQ on L is P'O', the segment intercepted between the feet of the perpendiculars PP' and QQ'- Now P'Q' = PO cos (PQ . P'Q'). .". the projection of any segment on a given line is the segment multiplied by the cosine of the angle which it makes with the given line. From left to right being considered as the + direction along L, the segment PO lies in the ist Q., as may readily be seen by considering P, the point frotn which we read the segment, as being the centre of a circle through O. Similarly OP lies in the 3rd O., and hence the projection of PO on L is + while that of OP is -. When PQ is _L to L, its projection on L is zero, and when II to L this projection is PO itself. Results obtained through orthogonal projection are univer- sally true for all angles, but the greatest care must be •:m 1 84 syntiiktk: (;kometry. excrrlscfl with regard to the concerned. sij,Mis of anguhir function! Kx. AX ;iiul OY are fixed lines at ri^ht nnjrles, and A(2 any line and P any point. IS Required to find the ±PQ in terms of AX, I'X, and the .A. Take P() as the positive direction, and X project the closed figure POAXI' on the ^ line of PO. Then P''-^'Q + P>-QA + pr.AX + pr.XP=o. (i68^ Now pr.PQ is PQ, and pr.OA^o; AX lies in ist (l, and XP m the 3rd Q. '" Moreover .(AX . J>(2), /..., the rotation which brings AX to I Q in direction is --_xN, and its cosine is +. cos/.(AX.PO)^+sinA. Also, pr.XP is -XPcos^XP(2= -XPcosA. PO = XPcosA AXsinA. .SIGNS OF THE SKCiMKNTS OF DIVIDED LINES AND ANGLES. 230°. AOB is a given angle and .A(JH= -.BOA. Let OP divide the -AOJB internally, and OO divide it ex- o ternally into parts denoted respectively a\ ^^ "' ^' ^^^ "''^ ^"V-^J"?'\ ^^ " ^^ ^^^ ^^^^ and ^ the ^POB, a L 1 and /3 are both positive. But if we I .M^^n \ > ^'''■''^ - ^""^ ^'' '^'^-AOQ, and /S =^i20B, and a and // have contrarv signs On the other hand, if a is ^AOP and Ahe ^llOP, a and 8 have contrary signs, while replacing P bv O gives a' and 3' with like signs. The choice between these usages must depend upon con- venience ; and as it is more symmetrical with a two-letter (IKOMKTklC KXTKNSrONS. 185 notation to write AOP, HOP, A()(), HO(), than AOP, POM, etc., we adopt the convention that internal division' of an' an-le -ivcs segments with opposite signs, while external division gives segments with like signs. In like manner the internal division of the segment AIJ gives parts AP, MP having unlike signs, while external divi- sion gives i)arts Af), VA) having like signs. D^/.~A set of points on a line is called a nnfj^r^ and the line is called its a.v/s. By connecting the points of the range with any j)oint not on its axis we obtain a corresponding pencil. (203", Def ) Cor. To any range corresponds a pencil for every vertex, and to any vertex corresponds a range for every axis, the axis being a transversal to the rays of the pencil. If the vertex is on the axis the rays are coincident ; and if the axis passes through the vertex the points are coincident. 231". BY is any line dividing the angle B, and CR, AP arc I)crpendiciilars upon BY. B Then AAPY«ACRY, AP is ABsinABY, CR is BCsinCBY, AY^AB sinABY / ^^ CY CB'sinCBY* ^^C Therefore a line through the vertex of a tri- ^^i" angle divides the base into segments which are proportional to the products of each conterminous side multiplied by the sine of the corresponding segment of the vertical angle. Cor. I. Let BY bisect lB, then ^^ = i:". YC a and and AY=^(/;-AY), and AY = A . "' a + c Thence YC = ha a + c *-: ill W 111 fif^ Iff I ill ! I J 3 i86 Which are the divides the base SYNTHETIC GEOMKTRY. segments into which the bisector of the lB AC. Cor. 2. In the AABY, HY-=:AE-'+AY^'-2AB.Ay, cos A. Hut cos A /^ + c- (2lf) (V (217^-), and AY = be whence by reduction which is the square of the length of the bisector. Cor. 3. When AY = CY, BY is a median, and AB^sinYBC CB sinABY" • '. a mr n > a triangle divides the angle through which it passes 1 rts whose sines are reciprocally as the con- terminous sides. 232°. In any range, when we consider both sign and mag- nitude, the sum AB + BC + CD + DE + EA =0, however the points may be arranged. For, since we start from A and return to A, the translation in a + direction must be equal to that in a - direction. That this holds for any number of points is readily se^n Also, in any pencil, when we consider both sign and mag- nitude, the sum -iAOB+z.BOC + _COD + .LDOA=:o. For we start from the ray OA and end with the ray OA and hence the rotation in a + direction is equal to that in a - direction. RANGES AND PENCILS OF FOUR. 133 . Let A, B, C, P be a range of four, then AI5.CP + BC.AP + CA.BP^o. .B GKOMETKfC EXTENSIONS. Proof.— AP = AC + CP, and BP = BC + CP. .•. the expression becomes BC(AC + CA)+(AB + BC + CA)CP, and each of the brackets is zero (232°). .-. etc. 187 234°. Let O . ABCP be a pencil of four. Then sinA0B.sinC0P + sinBOC.sinA0P + sinCOA.sinB0P=o. /"r^^Z— AAOB = iOA. OBsinAOH, also AAOB=UB.;^, where p is the common altitude to all the trian.srles. AB./ = OA.OB.sinAOB. Similarly, (ZV.p=OQ. OP. sin COP. AB.CP.j?j2=OA.OB.OC.OPsinAOB.sin'coP Novy, p'^ and OA . OB . OC . OP appear in every homoloo;,us product, .-. (AB.CP + BC. AP + CA.BP);i'^ = OA . OB . OC . OP(sin AOB . sin COP + sin BOC . sin AOP + sin COA . sin BOP). But the bracket on the left is zero (233°), and OA. OB OC 01) is not zero, therefore the bracket on the right is zero. g ^ d. r.;^5^'/''°"' ^' ^^^ perpendiculars PA', PB', PC be drawn to OA, OB, and OC respectively. Then , sinAOP=:^P sinBOP=|^:|;, etc., :md^ putting these values for sin AOP, etc., in the relation of 234 , we have, after multiplying through by OP C'P.sinAOB + A'P.sinBOC + B'P.sinCOA-o Or, let L M, and N be any three concurrent lines, /, ///. „ the perpendiculars from any point P upon L, M, and N respectively, then ^ /sinMN + ;/;sinNL + //sinOT-o where MN denotes the angle between M and N, etc. K 'lii ' A' 1 88 SYNTHETIC (GEOMETRY. 236^ Ex. I. Le t four rays be disposed in the order OA, P OB, OC, OP, and let OP be perpendicular' to OA. Denote :.AOC by A, and _AOH by B. Then 234° becomes r.inBcosA4-sin(A-B)sinJ-sinAcosB=o, ^. ., , ^ ..°''' sin(A-B)=sinAcosB-cosAsinB. Sm^larly by writing the rays in the same order and making -BOP a n, and denoting _AOB by A and ^BOC by B we obtam sin(A+B)=sinAcosB + cosAsinB. Also, by writing the rays in the order OA, OP, OB OC and denoting _AOP by A and _BOC by B, we obtain ' ' (i) when i.AOB=~], cos (A - B)= cos A cos B + sin A sin B ; (2) when £jV0C = ~|, cos (A + B) = cos A cos B - sin A sin B ; which are the addition theorems for the sine and cosine. E^x. 2. ABC is a triangle and P is any point. Let PX, ^^^» P2 be perpendiculars upon BC, A CA, AB, and be denoted by P„, pj, p^. ' respectively. Draw AQ || to BC to meet PX in O Then (235°) X B c D PQsinA4-PYsinB + PZsinC=o Butif AU is±to BC,AD=/^sinC = QX. (PX - /> sin C) sin A + PY sin B + PZ sin C = 0, -( P.t sin A) = <^ sin A sin C. ^(P,^sinA) = ^-sinBsinA = rt;sinCsinB, ^XI'«sinA)=;''[rt^^sin2Asin2Bsin2c}. Hence the function of the perpendicular PaSinA + PjsinB + P.sinC IS constant for all positions of P. This constancy is an important element in the theory of /nVmcar co-or^uia^t's. or Similarly, GEOMETRIC EXTENSIONS. )A, ar I we C, 3 189 237". A, B, C being a range of three, and P any point not on the axis, AB.CP-+BC.AP-' + CA.BP- = -AB.BC.CA. Proo/.~Let PO be ± to AC. Then AO = AC + CO, BQ = BC + CQ, and the expression becomes (AB + BC + CA)(PON-CO-')+BC.CA(BC-AC) = BC. CA(BC + CA) = BC . CA . BA = -AB.BC.CA. Exercises. I. A number of stretched threads have their lower ends fixed to points lying in line on a table, and their other ends brought together at a point above the table. What is the character of the system of shadows on the tabic when (a) a point of light is placed at the same height above the table as the point of concurrence of the threads ? (b) when placed at a greater or less height 1 2. If a line rotates uniformly about a point while the point moves uniformly along the line, the point traces and the line envelopes a circle. 3- If a radius vector rotates uniformly and at the same time lengthens uniformly, obtain an idea of the curve traced by the distal end-point. 4- Divide an angle into two parts whose sines shall be in a given ratio. (Use 231;, Cor. 3.) 5- From a given angle cut off a part whose sine shall be to that of the whole angle in a given ratio. 6. Divide a given angle into two parts such that the product of their sines may be a given quantity. Under what condition is the solution impossible ? Write the following in their simplest form :— 7. sin(7r-^), sin(^ + /?), sin /, (^-l)K C0S(2. + ^), COs{2.-(^-J)},cos{^-(J + ,)), '- If it-*' » il I90 SYNTHETIC GEOMETRY. S. Make a table of the variation of the tangent of an angle m magnitude and sign. 9- OM and ON are two Hnes making the^MON = «, and PM and PN are perpendiculars upon OM and' ON respectively. Then OP sin« = MN. 10. A transversal makes angles A', B', C with the sides BC CA, A B of a triangle. Then sin A sin A' + sin B sin B' + sin C sin C'=o. 1 1. OA, OB, OC, OP being four rays of any length whatever AAOB.ACOP+ABOC.AAOP+ACOA.ABOP=o.' 12. If r be the radius of the incircle of a triangle, and r, be that of the excircle to side n, and if p^ be the altitude to the side (r, etc., ^ (sinA + sinB + sinC) sin A and " ^?A^ - sin A + sin B + sin C), ''i ^'2 fs Pi p, p^ r (Use 235°.) 13. The base AC of a triangle is trisected at M and N, then BN^'=i(3BC- + 6BA^-2AC^). SECTION II. CENTRE OF MEAN POSITION. 238°. A, B, C, D are any points in line, and perpendiculars ^ " AA'j BB', etc., are drawn to any fixed line L. Then there is, on the line, evidently some point. O, for which _______^__^^ AA'+BB' + CC'+DD' = 40N ; "a^ bH^j c' d' ■- ''ind ON is less th;m AA' and o-reater than DD'. The point O is called the centre of mean position , or simply the mean centre, of the system of points A, B, C, D. CENTR?: OF MEAN TOSITION, Again, if we take multiples of the perpendicul t> . V>V>\ etc., there is some point O, on the 191 ars, as rt.AA', , . , - , - axis of the points, for which Here again ON hes between A A' and DD'. O is then called the mean centre of the system of points for the system of multiples. />/- For a range of points with a system of multiples we define the mean centre by the equation 2:(.?.AO)=o, where 2(^ . AG) is a contraction for ^?.A0 + /;.B0 + 6. CO + ..., and the signs and magnitudes of the segments are both considered. The notion of the mean centre or centre of mean position has been introduced into Geometry from Statics, since a system of material points having their weights denoted by a h, c\ ..., and placed at A, B, C, ..., would "balance" about the mean centre O, if free to rotate about O under *>v^ action of gravity. -cuire ofuit The mean centre has th^-" —^ Perpendiculars from the "centre of grav^' " -- " <- + ^/3+^7=o, ^ passes through the centre of an excircle, that . for example, aa = d^ + cy. 239°. 77/ any range Exercises. Proof j''^^ ^° moves that the sum of fixed multiples of the ^•pendiculars upon it from any number of points is . ^ant, the line envelopes a circle whose radius is ^^2(^.AL)^ But, if O is tl. ^(^) .n centre of the vertices of a triangle, for equal pies, is the centroid. .an centre of the verticcb " .my regular polygon, for Ex. The 1 multiples, is the centre of its circumcircle. N I 192 SVNTI I K'J'IC (] EOMETRV. when the multiples are proportional to the opposite sides is the foot of the bisector of the vertical angle. 240^ LetA, B, C, ... be a system of points situated any- ^, ..., AM, im, CM, 1 I^ C, ... upon two lines L where in the plane, and let AL, I ..., denote perpendiculars and M. Then we define the mean centre of the system of points for a system of multiples as the point of intersection of L and M when 2:(''.AL) = o, and l(a.AM) = o. If N be any other line through this centre, i:(^.AN)=o. For, let A be one of the points Then, since L, M, X is a pencil of three and A any point, (2-11;°) AL . sin MON + AM . sin NOL+ AN . sin LOM =0 BL.sinMON + BM,sinNOL + BN.sinLOM=o,' also wj^' .^. lijst by a, the second by />, etc., and adding, "^^sinNOL+v(,;.AN)sinLOM=S.' 'I, by dff^nition, SECTION II. CENTRE OF MEAN POSITION ''''^''"' "^ • itever, 3^°. A, B, C, D are any points in line, and perp AA', BB', etc., are drawn t hen line L. Then there is.. ML, evidently some point. G. ML, AA' + BB' + CC'+L • • . "a^ eHJ c' D^ ■- and ON is less than A than DD'. The point O is called thecai/r^' of mean posih the me(w cenin\ of the system of points A, B, C. CENTRE OF MEAN POSITION. 193 242°. Theorem.- -The mean centre of the vertices of a triangle with multiples proportional to the opposite sides is the centre of the V incircle. ^® Proof.— IlcX^q L along one of the sides, as BC, and \Q\.p be the ± from A. Then A 'L{a.AV) = a.p ^^^ 2(.'?).OL = (rt + ^ + f).OL, ••• (239°) 0L = ^— =^=r; (153°, Ex. I) i.e., the mean centre is at the distance r from each side, and is the centre of the incircle. Cor. f. If one of the muUiples, as a, be taken negative. 0L = ■ap -A__ r'; ^,at '» (153°, Ex. 2) -a + o + c s-a vjj> / I.e., the mean centre is beyond L, and is at the distance r' from each side, or it is the centre of the excircle to the side a. Cor. 2. If any line be drawn through the centre of the in- circle of a triangle, and a, p, y be the perpendiculars from the vertices upon it, aa + dp + cy=o, and if the line passes through the centre of an excircle, that on the side a for example, aa = b^-\-cy. Exercises. I. If a line so moves that the sum of fixed multiples of the perpendiculars upon it from any number of points is constant, the line envelopes a circle whose radius is 2(^) The mean centre of the vertices of a triangle, for equal multiples, is the centroid. The mean centre of the vertices of any regular polygon, for equal multiples, is the centre of its circumcircle. N 2. "' Hi r Hi m 194 I SYNTHETIC GEOMETRY. and 243°. neorem.-li O be the mean centre of a system of Tthe Xr"" '''™'"P'"-"-y-d^Pen/ent point 2(«. AP=) = -(«. A0=) + v(^). 0P=. Pron/.-Ut O be the mean centre, P the independent point, and A any point of the A system Let L pass through O and be per- pendicular to OP, and let AA' be perpen- dicular to OP. Then Ap2=.AO'-'+OP2-20P.OA', ^•AF'' = a.A0'' + a.0P^-20P a OA' Similarly '^. BF=^. BO^-f-^. 0P-20P.^. 'oB'; But S::o3:3::^5l-^"'-«^ Cor. In any regular polygon of « sides ith tlie sum of the ZnZ°" '"^ J°'"\°f ^"y P«i>" with the 'vertices is greater than the square on the join of the point with the mean centre of the polygon by the square on the circumradius. f or makmg the multiples all unity, 2(Ap2)=«r'+«Op2, ,^2(AP=) = OP2 + ;-=. Ex. Let a,i,che the sides of a triangle and „ « ,1, Jou,s of the vertices with the centroid. Tht' ("'{'e'^. V' S(AP-^)=2(A0=) + 30P=. 1st Let P be at A, p+,:^^,^„, , whence «H^= + .==3(„=+^+y).'^ '"^^^' E... If AECDEFGH be the vertices of a regular octa<.nn taken m order, AC= + AD=+AE'+AF= + AG==fP. CENTRE OF MEAN POSITION. 195 244°. Let O be the centre of the incircle of the AABC and let P coincide with A, B, and C in succession. 1st. bc'-\- ct^== z(a . AG-) + 2(.0AO2, 2nd. .7^2 +ca- = Z(a.AO-) + Z{a)BO\ 3rd. al>- + da'' = 2(rt . A0-) + 2(rt)C02. Now, multiply the ist by a, the 2nd by l>, the 3rd by c, and add, and we obtain, after dividing by (a + fi + c), ^{a. AO^) = adc. Cor. I. For any triangle, with O as the centre of the incircle the relation ^a.A?'') = s(a.A0')+7:{a)0F^ becomes ^a.A?'-)=adc+2s.0P% and, if O be the centre of an excircle on side a, for example, 2(r? . Ap2) = _ adc+2(s - rt)0P2, where a denotes that a alone is negative. Cor. 2. Let P be taken at the circumcentre, and let D be the distance between the circumcentre and the centre of the mcircle. Then AP = BP = CP = R. 2sR^=al>c + 2sD\ ^ :^ ^_^^ J°»\^Y, CZ, and through E," H,n», iM- . "^^ intersection of these ioinc: diaw DL to meet the altitude BH in O ^ ' D Y iflUc!' Jm l:^""^^^'^^'-^ '-^"d E is the centroid. Since ^^ IS II to BH, the tnangles YDE and BOE are similar BE = 2EY, ^ss% Cor.) , OE = 2DE, and as D and E are fixed points, O is a fixed point. • ■ the remam.ng altitudes pass through O. 249°. r/,.^;m.-Three concurrent lines perpendicular to thesides ofa trandeaf Y v 7 a; -j .1 •, t''="""-"'ar to RY27rv2 A ;.. ' ^ ^'""'"^^ ^^^ ^'^es so that BX^+CY2+AZ^=.CX-+AY^'+BZ-; OK COLLINEAKITY AND CONCUKRKNCK and, conversely, if three lines perpendicii triangle divide the sides in this manner, the lines are concurrent. Proof,— U\ OX, OY, 0/ be the lines. r Then IiX'-'-CX='=IJO'^-CO-,(i72',i) 2 Similarly C Y- - AY*J= C0-» - AO-, 199 A Y BX« + CY-"+AZ'^-CX^-AY--BZ-' = o. Conversely, let X, Y, Z divide the sides of the triangle in the nnnner stated, and let OX, OY, perpendiculars to BC and CA, oieet at O. Then OZ is ± to AB. Proof.— If possible let OZ' be _L to Ali, Then, by the theorem, BX=^ + CY- + AZ'=^ - CX- - AY- - BZ'-=o, and by hyp. BX-' + CY'- + AZ--CX-- AY-- I3Z-=o,' AZ'2-AZ^=BZ'2-BZ2. But these differences have opposite signs and cannot be equal unless each is zero. .-. Z' coincides with Z. I. Exercises. When three circles intersect two and two, the common chords are concurrent. Let S, Si, S^ be the circles, and A, B, C their centres. Then (113°) the chords arc perpendicular to the sides of the AABC at X, Y, and Z. And if r, r^, r., be the radii of the circles, BX2-CX2=r,2-r/,etc.,etc., and the criterion is satisfied. .'. the chords are concurrent. The perpendiculars to the sides of a triangle at the points of contact of the escribed circles are concurrent. When three circles touch two and two the three common tangents are concurrent. If perpendiculars from the vertices of one triangle on the sides of another be concurrent, then the perpendiculars fh :) ■i '' 1 \m ■ i ii >!'] 'i^.- 'M wM ^r iH it'i^H IH it H t,f i ' '^^1 ^ '^^H 4 - ^^^^1 ^ ^^1 li . ^^^1 1^ ■ 200 SYNTHETIC GEOMETRY. 6. Two perpendiculars at points of contact rf^!'.- , concurrent with a perpendicuh rnT f " "^ the incircle. P^'P«"d.cular at a point of contact of .He sMesinto'^^tt thict C:^-- r^^^^^ '"^^ ^'^'''^ (a) BX^Y.AZ_ CX.AYJBZ""^' (^) sinJ^AX^sinCB Y . sin ACZ sin CAX . sin AB YT^hHIcZ " '' Proof of {a).~ On the axis of X Y 7 H,-., .u B °' ^' ^' ^ ^'^w the perpendicu- lars AP, BQ, CR. On account of similar As ^=BQ CY^cR AZ \v CX CR' AY-AP' BZ = BQ' BX.CY.AZ Proof of{b\ Similarly, ex. AY. BZ BX^ ABAX^ BA sin BAX CX ACAX CAsiFcAX' sinBAX^CA BX sin CAX BA'CX' s4nCBY^AB CY sin ACZ BC smABY -^•-^' ^^is'^ = 1. CB"AY' sinBCZ^AC smBAX^^inCBY.sinACZ BX CY A7 sm CAX. sin ABYT^n-BCZ" ' ' AZ BZ' cx^ayTbz ~ ' V-^.«'. ine preceding functions vvbiVh nv-n • • s icixns Which arc criteria of colhnearitv OF COLLINEARITV AND CONCURRENCE. 201 will be denoted by the symbols VCX>^^"^(si,TCAXr'^'P^'''^^^y- It is readily seen that three points on the sides of a triangle can be collinear only when an even number of sides or angles (2 or o) are divided internally, and from 230° it is evident that the sign of the product is + in these two cases. Hence, in applying these criteria, the signs may be dis- regarded, as the Hnal sign of the product is determined by the number of sides or angles divided internally. The converses of these criteria are readily proved, and the proofs are left as an exercise to the reader. Ex. If perpendiculars be drawn to the sides of a triangle from any point in its circumcircle, the feet of the perpendicu- lars are collinear. X, Y, Z are the feet of the perpendicu- lars. If X falls between B and C, Z.OBC is < a ~|, and therefore ^OAC is > a ~], and Y divides AC externally ; .'. it is a case of collinearity. Now, BX = Ol3cosOBC, ''^nd AY=OAcosOAC. But, neglecting sign, cos OBC = cos OAC, BX OB AY CY OC AZ_OA OC and similarly, BZ OA' OC OB' cx VCX/ '' ^"^ '^' ^' ^ ^^'^ collinear. /Ay:_The line of collinearity of X, Y, Z is known as bmison's Ime for the point O." !.<■ ''ii n •m 251°. T/ieonv/i.~When three lines through the vertices of a triangle are concurrent, they divide the angles into parts 202 SYNTHETIC GEOMETRY. which fulfil the relation (n) sin BAX^sinc^B Y . sin ACZ relaU':^ cl.v,de the opposite sides into parts which r.m the (^) BX.CY.AZ__ 7- a CX.AY BZ '• ^ ^t-, OQ, OR be perpendiculars on the sides. Then 'sinBAX\ /smBAX\ \sinCAXV==~-'» sinBAX^OR sin CAX OQ' sin_CBY_OP sinABY~"OR' sinACZ^OQ sin BCZ OP' A QY .". multiplying, c^vidS:;i; "^"'"-^ ^™'" '"^ "=- -gles being penltrTuit "/r ' '"' ' '^' '^^ -" ^F be pe^- Then, from similar As BEX and CFX, BX^ BE^AB sin BAX CX CF ACliSCAX' Similarly, ^^"^fii^^Y A2_CAsinACZ . . m„U,ply,„g, (5X^=_,_^_.^_^^^^ The negative sio-n v«c,,u /• ^^.^.rt'. internnlV "" '"""^ '■°'" "'" 'hree sides being divided Jt is readily seen thnf ti.,-o<. (230°) vertices of a .'ianlle Is d, /°"""™"' ""^^ ""-""eh the -d of sides inter^ ■;:;•/;'': tTe "r T'""" "' "■'"^'^^ P'-ocIuct is accordingly'^egative ""^ ''°'" "^ "'^ Hence, in applying the critpi-la ,!,« • be neglected. ™' "'^ ''S"« of the ratios may OF COLLINEARITY AND CONCURRENCE. 203 The remarkable relation existing between the criteria for colhneanty of points and concurrence of lines will r ce v^n explanation under the subject of Reciprocal Polars 252° Exercises. r. Equilateral triangles ABC, BCA', CAB' are de- scribed upon the sides AB, BC, CA of any triangle. Then the joins AA', BB', CC are concurrent. Proo/.—S'mce AC'=AB, AB'=AC and • • But Similarly, z.CAC'=4BAB', ACAC=AB'AB, and ^AC'C = MBB' i!llACZ=!i!lACC'_AC'_AB sinABY sin AC'C AC~AC" ^^^^°^ sinBAX^BC sin CBY CA CB' 2. 4. sinBCZ BA' sinCAX^ /sinBAX\ _ VsinCAX/ '' _^and hence the joins AA', BB', CC are concurrent. The joins of the vertices of a A with the points of con- tact ot the incircle are concurrent. The joins of the vertices of a A with the points of contact of an escribed are concurrent. ABC is a A, right-angled at B, CD is = and ± to CB and AE is = and ± to AB. Then EC and AD inter- sect on the altitude from B. 5- The internal bisectors of two angles of a A and the external bisector of the third angle intersect the opposite sides collineariy. 6. The external bisectors of the angles of a A intersect the opposite sides collineariy. 7. The tangents to the circumcircle of a A, at the vertices of the A, intersect the opposite sides collineariy. T Tl ^' ^"^"'""^ '^ '^^ ^^'-^'^^^ °f ^ A, the lines hrough the point perpendicular to those joins intersect the opposite sides of the A collineariy. f m ? 1 f J ;>f \m 'Am « ] 1 ' \ 1 ii J k §■ 204 SYNTHETIC GEOMETRY. lo. II. ?nu '""" °' ••* ^ '■" ''"^ P°i"'' =0 'hat three of them connect with the opposite vertices concurrently fvit^h^'' "'^••«'"■->'':■"S three connect concurrently With the opposite vertices '' scrib^r T"' if ''"• ' '''' "'^" ^^^ As are all de- scribed internally upon the sides of the given A ' and A 3 C , and O is any point on L, A'O, B'O, and C O intersect the sides BC, CA, and AB col inea ly. -53 . 7y/.vm;^-Tvvo triangles which have their vertices connecting concurrently have their corresponding sides nter secting colhnearly. (Desargue's Theorem.) ?z ABC, A'B'C are two As having their vertices connect- ing concurrently at O, and their corresponding sides in- tersecting in X, Y, Z. To prove that X, Y, Z are col- linear. ■iy '^f^f-^-'^o the sides of I Y AA'B'C draw perpendiculars i AP, AP', BQ, BQ', CR, CR'. I Then, from similar As, CX CR' CY^CR' AY AP' AZ^AF BZ BQ' /BX\ AP'.BO'.CR' Vex/ ' ^ But AP. BQ. CR- AP' ^ si nu\A'B' AP sinAA'C" with similar expressions for the other ratios. OF COLLINEAKITY AND CONCURRENCE. 20$ Also, since AA', BB', CC are concurrent at O, they divide the angles A' B', C so that sin AA'B\ sin BB'C. sin CC'A'_ sin AA'C. sin BB'A'. sin CC'B'~ ^' ^Cx)"" '' ^"^^ '^' ^' ^ ^^'^ collinear. The converse of this theorem is readily proved, and will be left as an exercise to the reader. Ex. A', B', C are points upon the sides BC, CA, AB re- spectively of the AABC, and AA', BB', CC are concurrent in O. Then 1. AB and A'B', BC and B'C, CA and CA' meet in three points Z, X, Y, which are collinear. 2. The lines AX, BY, CZ form a triangle with vertices A", B , C", such that AA", BB", CC" are concurrent in O. • OF RECTILINEAR FIGURES IN PERSPECTIVE. 254°. Z'^Z-AB and A'B' are two segments and AA' and BB' meet in O. A n Then the segments AB and A'B' are said to be in perspectwe at O, which is called their cetttre of perspective. The term perspective is introduced from Optics, because an eye placed at O would see A' coinciding with A and B' with B, and the segment A'B' coinciding with AB. By an extension of this idea O' is also a centre of perspective of AB and B'A'. O is then the external centre of perspective and O' is the internal centre. Def.—Two rectilinear figures of the same number of sides are in perspective when every two corresponding sides have the same centre of perspective. '■'i h 'ifl S-f! d^H " 1 i^^l f.' •^ '1 ■ I,,* ^^^1 'i s 1 ■ i ^^1 , 1*, ^^^H '■' ) ^^^1 r ■ t ■ ^^H v{ ^^H ^1 ' »e ^^k jr* ' ^^^1 ."! H \m 206 Cor. I. I roctil (h mm (ho prcrrcliinr ,1 tOMl/r/< '"<'•»«• (i^nii OS of (I "«^«'<"erspecti\e, ^^c SIN points may 1 con- "'■'". '^^"■'•s <"■ In-wnKles h 'r '^^^""ccted i„ fonr di/Vc VIZ. .^'^c,A'irc;Anr.A . , lent ways '^•'"K^ the same centre of of ''"'lose foil perspective, which '^'^'AH'CA'MC; A'nc '• P^"'s of conjugate triangles d centres of perspective of "itorsect in it^tcinnne A Aire, «n pairs, three \, ^' / , til SIX points ; these poi„( ^ sides of the our axes s are v; /' I Thi th )ein.; internal points, tht <'entres r!" '"'"'''^ ^^^"''-^^^ and tl t^vo trian.i^Ies tal; lice X' en ^ segments ,>f .^hich '■"/orsect.ons which dot they ar ^'•"iinc them, and .^•■vcn in the followin,. t.d)le L''"' ''"''"'' "^ ^^^''^'^i^^^-l ivc arc ^(M\r. X n In 'l^ri'K.MiM,.,, iiv z X' nc~irc' CA - C'A' AH-A'ir I5c'-irc CA'-C'A Air-A'n Pi (^'kntkk or lOSPKCTIVir •, I5C'-I}'C CA'-C'A Air~A'M () 'f^> nc ~ n CA - C'A' or INVKKsrON AND INVF-; KSK KHiUHKs. 2 And the six points lie on tl o; AV/, X'V'/ x'\ >c 'our lines thti? >^', XY7/ I- The diandric fonncd KXKKCISI-.S ex ''y j nvrlvs of.uiy liinnj^lc i oininj; the rcnti cs of tlic thi cc 2. Tiio three rhoids of an|;le fo ^ Ml per.spec tive with it. <<»nt.irt of the cxriirles of .3- 'Ihc tanj^ents to tl nn.i tn.ingic in perspective with tl riny tri- le ori|;Jnal vertices form a t H" nrcnnicircle of a triangle at the tl n iiiiK'c in pcrspcc irec live with the original. SICCTION IV. OF INVERSION AND INVICKSF. FIGURKS. ciXh^^;;''r '"'"^^ "' "^"''^"' -i^"" - ^'-'^t-iinc of a respect to the circle. ''"'^' '"'"'''"' '^"'"^'^ ^'t'' Thus 1> and O are inverse points if R being the radius. I c~p~7b 2r The 0S is the cm/^' ofim>crsion ^ ^ "«• the inverting 0, and C is the centre of inversion. Cor. From the definition :— >. An in,lonni,c nun.bcr .,f ,„i,s of inverse poin.s mny lie on 2. An i,uleln,i,e nun.ber .,f rir,lc,, may have the san.e two pomts as mvcrse points. 3. 1'"''' points of a pair „f inverse points lie „po„ the snn.e side of the centre of inversion ml ii 20i^ 5- I' and O SVNTIIKTIC (;i:()MKTI so that (I H' inverse of any point at inliniiy, aie l)oth nci;ativc. lUit K^ k„:„ .. , " '"^'vi.L.n Hence, in onlcr th^it the riirl^ „r ■ oaoh pair or points „,„,'T^ of "vers.on „,ay be real, OI- INVKKSroN AND INVFJ nalIy Inltcr. ' '"' '"^y ^"^'"I'^'-Iine (,f the I. V and () arc inverse to 0S. ''"lien CI'. CO. .('I'-! CT is tangent t( 0S'. A , , ('7^.'. Cor. 3) ^"d .-. S cuts S orthogonally since the radius of S is perpcn- diadar to the radins of S' at its end-,W 2. Conversely, let S' c ,,1 S orthooc,naliv. Then lCTC •s^. n. and therc.,re CT i^t^^^^^^ ■^"fl P and O are inverse points to 0S, Cor. ,. A through a pair of points inverse to one another ^vuh respect to two 0s cuts both orthogonally. Cor. 2. A which cnts two 0s orthogonally deterndncs on then- common centre-line a pair of points whicl are i to one another with respect to "both 0s. Cor. 3. If the 0S cuts the 0s S' and S" orthogonally the ..ngcMUs frou. the centre of S to the 0s S'nnd S'' re nuh!^ o anu tncreforc equal. o n 1 I I' t If-I ■^ 111 - f\ n 2IO SVNTHKTir C;i.:f)Mi;TKY. LMv^n'n.^ ®, ''■■'""*'' '" ^""■"' "" "«^ '■•"'"■••>' ••""■^ of two K ven 0s, a,ul ,utl,„K o„c <,f ,1,™, ..rthogcnalK., cuts the (itlicr ortlio);c iially also. '' " ;;vo n.e., PO.US ,K.vo a constant „u.;;:;;;;;,;ie;;:r^.,;: tuo fixed points as inverse points. "^ Cor. 2. When ]) comes to A and H wc obtain ()!) OA Qir "cnce DA and 1)15 are the bisectors of the ^PDQ .nd tl>e se.,n.ents VU and P.O subtend ec,ual angles ar ) ' Hence the locus of a point at whicli two adjacent Ve^^ments c^ s H,e subtend cc.ua, angles is i circl^^^C ei o h r" ""'-^"'"'^ ''' '''' ^^^'"^"^^ and having then other end-pouits as inverse points. OF INVKKSION AND INVKKSK I.„;ui<|.:s. 21 1 il'DI =^ll|)()', and i.lil)l'==.^|ii)() i.l'l)r'=.-a)l)()'. ~' -T the soKme,,,, VI" ami y,V sub.cn.l ,..|„al angles a, I> n,.e he Iocs of a ,.oi„. a, «„i„, ,J,, „„ni|ja , 'sck- "- <-'»'ii"iiils as pairs of nivcrsc poinls. Cor. 4. Since Al> : A(i=PIi ,• iso ,,. ••• 1' ■■>"<1 y ' nn,i d" . v-:)/ , ^x. i; centre lino re • If • ' '^"^' ^''^ common c Tn 1 u '" ^^^ '^^J^'^'^f' Points O and ()' 5- To describe a circle tn pi^o thr-n-i. • " - ' .wo given circles orth':s;nIn;: ' " """' ""'"' '"" "" & Wt: & It i !H If h ^ 212 SVNTlrKTIC GKOMF.TRV. ' |^""ccs fro,,, two fixed „s is ^,ivcn. 7. io (,„c| a point „,„,„ ., j,ive„ |i„c fr„„, ,vl,i,i, ,l,e ,„.„ „f noh!r^ -"^-""e fipirc is the inverse of anotl.er when cverv po,nt on one (,,M„e has its inverse .,,,on the other fi^ine ccnt.c of .aversion ,s no, on the nj,n„e to l,e inverted. s Let O be the centre of inversion and S be the drrln f. k nvertcd • nnM ]nt A' n' n> 1 .1. . circle to be nvmea and let A, B, C be the inverses of A, B, C resoec tively. 1 o prove that the ^B'C'A'=:~J. ^ AOA'C'« AOCA, and AOH'C^ AOCD ^^^ ^ .-. (i) .OCA'=^OAC, and (.) .OC'B'= 2^)1,0 And ^B'C'A-.OC'A' - OCB-.OAC - 'oBc =-AC]i=n. since ACB is in a semicircle OF INVKRSION AND INVICRSK FIGUKIiS. 2f3 ta.!^"nts: "'"' '""' "" " '" '"^'"^^^^'°" "^ ^-"-- direct Cor. 2. •.•OI)'.On = OT.Or.-^^^' l'I^ = Ka OD'^ where R is the radius of the circle oHnJcLn;^ ; . the centre of a circle and the centre of its inverse are not inverse j)()Mits, unless OD-dt /. i ";^'''C nie not inversion is at 00. ' '''" ""'"^^ ^''^ ^^'^^'^ ^''" Cor. 3. When the circle to be inverted cuts the circle of ~n, us inverse cuts the circle ^ (256", Cor. 5) 261". 77i,or,m. A circle which passes through the centre of inversion inverts into a line. Let O be the centre of inversion and ^S the circle to be inverted, and let V and V be inverses of Q and ()'. Proof.- Since OP. 00 -(M". ()()' = R2 .-. OP:OP' = OQ';OQ, ' and the trian^dcs OPP' and OO'O ^j are similar, and z.OPP' = _0()'()ln sinro OOY^ • • semicircle. And as this is tr^ howj;^ OP' b^d?.:, 'l;:> ;s a line X to OP, the common centre-line of the ^ de L inversion and the circle to be inverted. \ ^/ Cor. I Since inversion is a reciprocal process th^ inverse of a line is a circle through the centre rf in and so situated that the line is \ to th^ " """ line of the two circles. '^' '"'""^"" ^^"^'•^- Cor. 2. Let I be the circle of inversion, and let FT ind T>T' be tangents to circles I and S respectively ThL PT^=OP^-OT^=OP^-OP%(^ = Sp.P;r=PT'^ PT=PT', ■<' ^ "■ ^ .'. when a circle inverts into 1 lino „-;ti. - c.>c,e. ..e line . ..e .dical Xr r^l' STd^sr^S ■;u: -I i. I l\ h I 'I m m versa. 2 '4 SYNTHETIC GEOMETRY. and'cuN ,h! ''. "'■■'^'^P''^^*' "'■•""gh the centre of inversion chord "' '"™"'°"' ''^ '"^-^'^^ i^ ">«- common Cor. 4. A centre-line is its own inverse, its te^e^r Hn:i.:t ""^^ ^^ ^"^^^^^°" ^^ ^ ^^-^^-'^^ 262°. A circle which cuts the circle of inversion orthogon- — ally inverts into itself. Since circle S cuts circle I orthogon- ally OT is a tangent to S, and hence OP.OO = OT2, .-. P inverts into Q^and into P, and the arc TQV inverts into TPV and vice q.e.d. Cor. Since I cuts S orthogonally, it is evident that I inverts into Itself with respect to S. '"vens 263°. A circle, its inverse, and the circle of inversion have tl-- a common radical axis. Let I be the circle of inver- sion, and let the circle S' be Q--^_^^ , , I ,- the inverse of S. The tangents TT' and VV meet at O (260°, Cor. i), and the middle poini of TT' il on \7 T l""^""'^ P'^'"'"' ^' ifyjiiii ui 11 is on the radical ax s of S nnri c and the crcle with centre at D and radius DT cms S and S= or o^onaliy. But this circle also cuts circle I o^irnaij (258 )• . . D IS on the radical axes of I, S and S' radtl'ats o/Srd^;^ '''""' "'^'^ -^-' ^V, is on the .tif;:.^r^" ''.,!-"'• ^'^' »-- ^ ^•"-n radical -7 ~, w...^ axis passing through D and D'. q.e.d. OF INVERSION AND INVERSE FIGURES. 215 /J.v;/^r^..-This is proved more simply In- supposing one must cut the circle of inversion in the same noints and the common chord is the common radical axis ' Jof r::;;;" "^^^ ^^---ersection fbllows fVom the cirl's^s'IoT^•~?' ""''' '' intersection of two lines or ciicles is not changed in magnitude by inversion. A-et O be the centre of in- version, and let P be the point of intersection of two circles S and S', and Q its inverse. and Similarly Take R and T points near P, o and let U and V be their in- verses. Then OU.OR = OQ.OP = OV.OT = R2 AOQU^AORP, ^OQU=^ORP = ^RPX-^ROP ^OQV=^TPX-z.TOP, lUQV=lRPT~lROT. But at the limit when R and T come tn P fi.« i ,. the chords RP .,nri PT K , '^^ """^^^ between M c° n / o"^ ^ ^^'°"'^' '^^ ^"^'^ between the circles (ii5,Def I. io9°,Def.i). And, since ^ROT then vanishes we have ultimately z.UQV=^RPT vanishes, Therefore S and S', and their inverses Z and Z\ intersect .t the same angle. "' Cor. I. If two circles or a line and a circle .o„ri, another their inverses also touch one another! " Cor. 2 If a circle inverts into a line, its centre-lines invert into circles havmg that line as a common diameter F^r smce the circle cuts its centre-lines orthogonally, their in-' lino " Z or.li.^^„ally. but tne centre-line is the only line cutting a circle orthogonally. ^ hV'i ii. >,hi ,.lli 1 2l6 SYNTHETIC GKOMETRV. EXERCISFS. '•''!';: iLv:rer"''°^^"™''''^-™".'ewi.h.. aspect .0 ' "Th„2:„:! ^^'^-""■"^^°" °^ ^ «'-" ci..c,e ,.,3 , „,. 4. A d,Ce cues two ci.Ces ort./og™. ''^l^rir"- .nto two chclcs and the,,- co°n„ onccm •" ine ''''"" 5. Three crcles cut each other ortho^ona^v f,' k • -^..oh„es.the,rh,tersect,o„'ir[Hel':::o^\: -05°. The two following ovTinnI«o ^ • ^*viii^ cxciuipJes are important C.C,:- .^hfcir::: it ::;;,::::arr; ^^"'■-"-- -" -^ which connect concurrently onClZr, :,■ ^ ''"""^ S and S' are the two given circles and '^ a circle cutting D them orthogonally, ^'ivert S and S' and their common centreline with re- vvhich cuts S and S' orthogonally nnH h'^^^' ^° "" '^"''^^^ point O on Z S and S^T. V '' '^' "^"^'"^ ^^ some centre-line into a chct th o"!; T ''^"^^^^^^^' '^^ ^^^-r tonally, /., il ,i"lez '' "" ""'"^ ' ^"^ S' -^ho- BB^;^r^'r-:r,::t;^:^^^^^^ f OF INVJiKSION AND IN ICKSIi FIGURES. 2i; T, and by the ex- circle to the side ^ at T'. Take CH = CAand CI) = CB, and join DH and HA. From the sym- metry of the fig- ure it is evident that HI) touches of AB^d Ac' ' T," 'r .^" ^ ••'"' "" ""= ">« '"'"'"<= PO""s or Ali and AC, and let EF cut HA in G From 135°, Ex. i, AT = BT' = /-^ fr'" X ET = ET' = |(.,-'/.). But, smce EF bisects HA, EG = lBU = l(a~/;) •*• . ET = ET' = EG, and the cn-cle with E as centre and E(; as radius cuts I and S orUaogonally, and, with respect to this circle, the r de" I and S mvert into themselves. Now, PF;HC = DF:DC = BC-CF:BC, ^^ 2a EP = EF-PF = ^-<^ + 2a^ EP.EF = («-^ + ^'^;.)^ = ^(„_,).^j,(., /.P inverts into F, and the iinc HD into the circle through t and F, and by symmetry, through the middle point of liC liiU this IS the n,ne-points circle (116°, Ex. 6). And since ihemselves ' ' ' '"™"" "' ' ""' ""' -- ' ->" « And similarly, the nine-points circle touches the two remaining excircles. ™° I I, ill 2l8 SYNTHETIC GEOMETRY. SECTION V. I OF POLE AND POLAR, respect to the crcle of mversion, and the point is the M^ot cen't"if^nii:h?i:;^::r'-'"^^^- BuSnoe[ht7ott\:h^,S:ir::ah^ to 00 along any centre-line M,l 1 ! ""'™ "^y S° are polars of the centre And ' "'•' "^'"""^^ "'^'•^f™™ there ,•«; hnf .« r ^f'^/'^^ at infinity, thus assuming that mere is but one hne at infinity. at S,'; ^i.-'^,''" ^^^' °^ ""'^ P"'"' °" ">« <=!'•>:'<= is 'he tangent Cor. 3. The pole of any line lies on that centre-line of .h. polar crcle which is perpendicular to the former!ine Cor. 4. The pole of a centre-line of the polar circle lies tZ "' ""'"-""^ ""^'^ '^ Pe'-Pen^dicular to the Cor. 5. The angle between the DoInr<; nf f,.,« e^ to the angle subtended hy t^^'^ts'Tt t^e'Tda: OF POLE AND POLAR. 219 (266°, Def.) OP and OQ are centre-lines of the polar circle I and PE, ± to 00, is the polar of Q. To prove that QD, J. to OP, is the polar of P. Proo/.~The As ODQ and OEP are similar. 0E:0P = 0r3:00, and.-. OE.OO = OP.OdI But E and Q are inverse points with respect to circle I, P and D are inverse points, ^"^ •'• DQ is the polar of P. ^,^^^ /?C/:-Points so related in position that each lies upon the polar of the other are ^o;iju^a/c' points, and lines so related that^each passes through the pole of the other are rouju^ate Thus P and Q are conjugate points and L and M are con- jugate lines. Cor. I. If Q and, accordingly, its polar PV remain fixed while P moves along PE, L, which is the polar of P, will rotate about Q, becoming tangent to the circle when P comes to U or V, and cutting the circle when P passes without. Similarly, if Q moves along L, M will rotate about the point P. Cor. 2. As L will touch the circle at U and V, UV is the chord of contact for the point Q. .-. for any point without a circle its chord of contact is its polar. Cor. 3 For every position of P on the line M, its polar passes through O. ^ .-. coUinear points have their polars concurrent, and con- current hnes have their poles collinear, the point of concur- rence being the pole of the line of collinearity i! ; ; I i ' " i. 1 li fri ' n .1 m ii:. WW. I. 220 SYNTHETIC GEOMETRY. ' Exercises. (n) when P goes to oo along IVI ; (/^) when P goes to co along OD ; (c) when P moves along U V, whLt is the locus of D > From any pomt on a circle any number of chords are the point ' '^'"* ^'^"" '" ^'' "'' '^^ ^'-^"^^^"^ ^t On a tangent to a circle any number of points are taken show that all their polars with respect to the circl pa s through the point of contact. ^ 268\ rW;;..-The point of intersection of the polars of two pomts ,s the pole of the join of the points. ^ n^t""^ 1^?. P""^^'' ""^ ^ ^"^ «^ C P^ss through A. Then A hes on the polar of B, and therefore B hes on the polar of A (267^ For similar B • . c reasons C lies on the polar of A. .-. the polar of A passes through B and C and is their join. Cor. Ut two polygons ABCD... and a^c... be so situated ^ that a IS the pole of AB, ^ of BC c of CD, etc. Then, since the polars of a and d meet at B, B is the pole of a^ ; smiilarly C is the pole of /;^, etc the venices of one a JU'^ontsST.!::™:' etch ca!I '• '"' P°'" ""-^'^ "^'"S '"« 'ame in OF POLE AND POLAR. 221 Def. I.- -Polygons related as in the preceding corollary are polar reciprocals to one another. Def. 2.— When two polar reciprocal As become coincident, the resulting A is self-reciprocal or self-conjugaie, each vertex being the pole of the opposite side. Def, 3.— The centre of the with respect to which a A is self-reciprocal is the polar centre of the A, and the itself is the polar circle of the A 269°. The orthocentre of a triangle is its polar centre. Let ABC be a self-conjugate A Then A is the pole of BC, and B of ' ^^ ^ ^ AC, and C of AB. Let AX, _L to BC, and BY, _L to AC, meet in O. Then O is the ortho- <^entre. (88°, Def.) Now, as AX is ± to BC, and as A is the pole of BC, the polar centre lies on AX. For similar reasons it ^'^^ ^^^Y- (266°, Cor. 3) .*. O IS the polar centre of the AABC. q,e.d. Cor. I. With respect to the polar of the A, the on AO as diameter inverts into a line J_ to AG (261°). And as A and X are inverse points, this line passes through X ; therefore BC is the inverse of the on OA as diameter. Similarly, AC is the inverse of the on OB as diameter and AB of the on OC as diameter. ' Cor. 2. As the on AO inverts into BC, the point D is inverse to itself, and is on the polar of the A (256°, 5) .'. OD is the polar radius of the A Cor. 3. If O falls within the A, it is evident that the on OA as diameter will not cut CB. In this case the polar centre is real while the polar radius is imaginary. (257°, Cor.) Hence a A which has a real polar circle must be obtuse- angled. ;|i t i 222 SYNTHETIC GEOMETRY. I I i i Cor. 4. The on BC as diameter passes through Y since Y is a ~I- But B and Y are inverse points to the polar 0. .-. the polar cuts orthogonally the on BC as diameter. Similarly for the circles on CA and AB. ^ .-. the polar of a A cuts orthogonally the circles having the three sides as diameters. Cor. 5. The^AOZ = ^B,z.BOZ=^,andz.OAC=4C-|). And CX = OCsinAOZ = OCsinB,also =-^cosC where d is the diameter of the circumcircle (228°) to the triangles AOC or BOC or AOB or ABC, these being all equal. /,,/'o t x o- •, , ("^, Ex. 4) Similarly OA = ^cos A, OB=rt^cos B. But OX = OCcosB=-./cosBcosC, K''=OX. OA=-^2cosAcosBcosC. In order that the right-hand member may be +, one of the angles must be obtuse. Cor. 6. R2=OC.OZ = OC(OC + CZ) = OC2 + OC.CZ, and 0C= -rt' cos C, and CZ = ^ sin B = ^', (.38°) R2 = ^2(i_sin2c)_^^cosC "^ If O is within the triangle, rt?2< ^(^2 + ^2+^2) ^^^d R is HTinginary. Il : Exercises. I. If two triangles be polar reciprocals, the inverse of a side of one passes through a vertex of the other. A right-angled triangle has its right-angled vertex at the centre of a polar circle. What is its polar reciprocal ? In Fig. of 269°, if the polar circle cuts CY produced in C prove that CY = YC'. ' 2. OF POLE AND POLAR. 223 4. If P be any point, ABC a triangle, and A'B'C its polar reciprocal ^ Ith respect to a polar centre O, the per- pendiculars from O on the joins PA, PB and PC intersect the sides of A'B'C collinearly. 270°. Theorem.-\i two circles intersect orthogonally, the end-pomts of any diameter of either are conjugate points with respect to the other. Let the circles S and S' in- tersect orthogonally, and let PQ be a diameter of circle S'. Then P' is inverse to P, and P'Q is ± to CP. • •. P'Q is the polar of P with respect to circle S. .*. Q lies on the polar of P, and hence P lies on the polar of O, and P and Q are conjugate points (267° and Def). Cor. I. PQ2=CP2+CQ2-2CP.CP' (172C2) = CP2+CO=^-2R2 = CP2-R2'4.CQ2_R2 = T2^T'2^ where T and T are tangents from P and Q to the circle S. .-. the square on the join of two conjugate points is equal to the sum of the squares on the tangents from these points to the polar circle. Cor. 2. If a circle be orthogonal to any number of other circles, the end-points of any diameter of the t^rst are conju- gate points with respect to all the others. And when two points are conjugate to a number of circles the polars of either point with respect to all the circles pass through the other point. 271'. Theorem.-^h^ distances of any two points from a polar centre are proportional to the distances of each point from the polar of the other with respect to that centre. (Salmon) i- i'd m\ 224 SYNTIIKTIC GKOMETRY. NN' is the polar of P and MM' is the polar of Q. Then PO:QO = PM:ON. Profl/.~LQi O/// be || to MM' and O// bell to NN'. Then OP'. OP = OQ'.OQ OP^OQ'^Mw OQ OP' N//* Put the triangles OPw and OQ// are similar, OQ On Nn ^Pw+ Mw_PM (>+N;/~ON* ^•''•'^• Cor. I. A, B are any two points and L and M their polars and.P the ponit of contact of any tangent N. ' AX and BY are ± upon N, and PH and PK are ± upon L and M respectively. Then |;^=!^and^X^AO PK R PH R' BY.AX_AO.BO , PK7ph K^ " ^ ' ^ constant, AX.BY = >('.PH.PK. If A and B are on the circle, L and M become tangents havmg A and B as points of contact, and AO--BO = R AX.BY = PH.PK. (See 211°, Ex. I) Exercises. ;ir r. If P and Q be the end-points of any diameter of the poh circle of the AABC, the chords of contact of the point P with respect to the circles on AB, BC, and CA as diameters all pass through O. 2. Two polar reciprocal triangles' have their corresponding vertices joined. Of what points are these joins the polars } Ml OF POLE AND POLAR. 225 3. A, \\, C are the vertices of a triangle and L, M, N the corresponding sides of its reciprocal polar. If T be a tangent at any point P, and AT is J_ to T, etc., A T.B T. CT _ AO . BO . C0_ PL . P M . I'N R3 ^ constant. If A, \^^ C are on the circle, AT.BT.CT = PL.PM.PN. 4. In Ex. 3, if A', B', C be the vertices of the polar reciprocal, A'T . B'T. C'T^ A^OJJ'O . CO AT.BT.CT R-J The right-hand expression is independent of the position ofT. 5. If ABC, A'B'C be polar reciprocal triangles whose sides are respectively L, M, N and L', M', N', and if AM' is the ± from A to M', etc., AM'. BN'. CL' = AN'. BL'. CM', and A'M.B'N.C'L = A'N.B'L.C'M. 272°. Zy/^^r^w.— Triangles which are polar reciprocals to one another are in perspective. ^p- Let ABC and A'B'C be polar recipro- cals. Let AP, AP' be perpendiculars on ^K^-..^^~\ — ^B . A'B' and A'C, BQ and BQ' be perpen- diculars on E'C and B'A', etc. Then (271°) ^=^, I^Q'=I^O ^ ' ^ BQ BO' CR CO' ' AP'. BQ'. CR' _ AP.BQ.CR ^' But AP' = AA'sinAA'F, AP = AA'sin AA'P', AP'^sinAA'P' AP sin AA'P' and similarly for the other ratios. Hence AA', BB', CC divide the angles at A, B, and C, so as to fulfil the criterion of 251°. .'. A A', BB', and CC are concurrent, ai in perspective. igl( are P (254°, Cor. 2) t'J s i;| 22G SVNTllKTR" (JEOMKTKV. ill SI'CTION VI. <>!■ TIIK KAUICAI, AXIS. .../..., I^ct a system of c-.A-drdes S S q ~M>oin,sPa,;^aandifi^'b;r:i;Sc::;: Then the centres of all the circle, of the r^-svsfom r L and hnv^^ tvt'at r .l • '•/'•-s\stcm lie 1- and ha^e MAI for then- common radical axis. on or 227 y THK RADICAL AXIS. Ilenrc from any point C in M'M the tangents to ill rho cnclcs are equal to one another. ^ '" ^'^.^ Let CT be one of these tangents. The cirrlp 7 - M, r centre Tnrl m^ ^„ j- <>^"i3. i iic cncie /. v»itn C as sirs s^r Metres ';;%^-''-^^^'-'^--«>-. with cen.es l,i„, on M'M suet ft 'eL^' one r.h^sS: c..^ orthogonally every one of ,1k ../.-circles. ' bmce the centre of any circle of this new system is oblaineH by drawng a tangent fron, any one of the circTs as S of the ..A-speces, ,o meet M.M', it follows that .o drdc of thts new systetn can have its centre lying between P and Q A r approaches P the dependent circle Z contract nmn'it be comes the po nt-circlo P «,h«.. -r ^ °^" Hence P Zn r '°"''' '° coincidence with P. Mcnce 1 and O are hm.tmg forms of the circles havin- c lied I v'" f . '^ '^" '"^'"^ ^>'^^^"^ ^^^ consequently called /w^^r^;^.r^,,,,^ ,,>,;,,^ contracted to /./.-circles From the way in which //.-circles are obtained we'see that from any pomt on L'L tangents to circles of the /./-system T: tT^^' "^r^ ^^'^^ L'L is the radical axis f th" /./.-circles. Thus the two systems of circles i^ ' .F^^ P have their radi- *:i 228 SYNTHETIC GEOMETRY. Cell axes perpendicular, and every circle of one system cuts every circle of the other system orthogonally. Hence P and Q are inverse points with respect to every circle of the /./.-system, and with respect to any circle of either system all the circles of the other system invert into themselves. If P and Q approach L, the .-c[rdes approach, and when P and Q coincide at L the circles of both species pass through a common point, and the two radical axes become the common tangents to the respec- tive systems. If this change is continued in the same direction, P and Q become imaginary, and two new limiting points appear on the line L'L, so that the former /^.-circles become r./>. -circles, and the former ^./.-circles become /./.-circles. Thus, in the systems under consideration, two limiting points are always real and two imaginary, except when they all become real by becoming coincident at L. Cor. I. As the r./.-circles and the /^.-circles cut each other orthogonally, the end-points of a diameter of any circle of one species are conjugate points with respect to every circle of the other species. But a circle of either species may be found to pass through any given point (259°, Ex. 5). .-. the polars of a given point with respect to all the circles of either species are concurrent. Cor. 2. Conversely, if the polars of a variable point P with respect to three circles are concurrent, the locus of the point IS a circle which cuts them all orthogonally. For let Q be the point of concurrence. Then P and O are conjugate points with respect to each of the circles. H^nce the circle on PQ as diameter cuts each of the circles ortho- gonally. ^^^^.^ Cor. 3. If a system of circles is cut orthogonally by two circles, the system is co-axal. For the centres of the cutting circles must be on the radical OF THE RADICAL AXIS. 229 axis of all of the other circles taken in pairs ; therefore they have a common radical axis. Cor. 4. If two circles cut two other circles orthogonally, the common centre-line of either pair is the radical axis of the other pair. Cor. 5. Two /^.-circles being given, a circle of any required magnitude can be found co-axal with them. But if the circles be of the ^./.-species no circle can be co-axal with them whose .diameter is less than the distance between the points. Exercises. 1. Given two circles of the /^.-species to find a circle with a given radius to be co-axal with them. 2. Given two circles of either species to find a circle to pass through a given point and be co-axal with them. 3. To find a point upon a given line or circle such that tan- gents from it to a given circle may be equal to its distance from a given point. 4. To find a point whose distances from two fixed points may be equal to tangents from it to two fixed circles. 275°. Theorem.~T\,^ difference of the squares on the tan- gents from any point to two circles is equal to twice the rectangle on the distance between the centres of the circles and the distance of the point from their radical axis. Let P be the point, S and S' the circles, radical axis, to AB. PX2- PT'^ wher and S' = PA2-PB2 e r, r are But, 273°, Def. ^^ Uh V:%\ 230 and SYNTHETIC GEOMETRY. =AB(AQ-QB)-AIi(AI-IB) This relation is fundanTenfafin'SrZy'f '.he radica/axt 4:nuare;Uat/:hf;p7;'::;r;, ^"r- -' '"^ tangents are not equal ' ™ *''" '^^"^^ ^^'^ ">e two 2c J'' "'"'' '"" ''^^"^ ^" ~'""- 'Agents to the Cor. 3. If p lies on the circle S', PT'=o and PT''= 2AB.pl, ' — ;r ,: : rie! TZTr '-' r -' °" °- ^^^^'^ - radical axis of the ctdes. '"'' "' '"^ ?<""' f™'" *« anaroi "w'th ;t* ""'pVi^AC tL^" ""^'"^ "'^°"^'' " Now, if P could .an,^.„e leav. th. circle we would have where PT" is the tangent from P ,0 'the circle S" •• PT2=P-P-PT"2 which IS impossible unless PT"=o on":rantV;rln^ "''"' ^^ "^^^ ^^- ^^e square radicaUxisorthis^^^::::^;;:;^:;^ 'feline is the """ '• '^T'^^'iT'^'T' ''\T ' '' ^ --^-^- Then PT2^ 2AB. PL 1) As PTJ varies as PL, P lies on a circle co-axal with S and S' circles are in a constant ratio is a circle co-axa! with the tuo II i OF THE RADICAL AXIS. 231 4. Exercises. In Cor. 5 what is the position of the locus for /&=o y&=xT >^=> I, /['negative.? ' ' What is the locus of a point whose distances from two fixed points are in a constant ratio ? P and Q are inverse points to the circle I, and a line through P cuts circle I in A and B. PQ is the internal or external bisector of the lAQII, according as P is within or without the circle. P, Q are the limiting points of the /^-circles S and S', and a tangent to S' at T cuts S a r in A and B. -*==-^b Then, considering P as a point-circle, tangents from any point on S to P and S' are in a constant ratio. .-. AP:AT = BP:BT, and PT is the external bi- sector of ^APB. If S' were enclosed by S, BT would be an internal bisector. 5. The points of contact of a common tangent to two /./.- circles subtend a right angle at either limiting point. * 276°. T/ieorem.~The radical axes of three circles taken in pairs are concurrent. Let Si, S^, S3 denote the circles, and let L be the radical axis of Si and S^, M of S, and S3, and N of S3 and S^. L and M meet at some point O, from which OTi = OT., and OT2=OT3, where OT^ is the tangent from O to S^, etc.,"' 0Ti = 0T3, andOison N, .'. L, M, and N are concurrent at O. De/.—The point of concurrence of the three radical axes of three circles taken in pairs is called the ra^ica/ centre of the circles. n^r^l'. !r ^^ ^" ^' ''"'.^ """? ^^ ^ ^^''"^ ''"'^^^ ^' t^^e common ^u 2 and S2, Z intersect on the radical t • \r^ and S, cis of S, ^32 SVNTHETlC GEOMETRY. Hence to find the radical tvI<= «f * -S. draw any two cwJz",lZT,T ""'" '' '""' The chords S, Z and S ? ' ^ ""* Siven circles. - .he Chords s:t^ndi!'z: -:::irj:- f ^-' -'^ coSi^-cLtr:;t:r::r'--Ssi^'v^- _ l^ee 249 , Ex. I.) »-or. 3. If a cn-cle touches two othf»r« fi.^ * points of contact .eet upon the^tr^^^S™" ''^ is ^tVeir'raditf;,™;: f-^'--. -*ogonaIly, its centre the radical centre to Tny o„"e of "hi"" '^ ''^ '='"^^"' ^-" ber'^^of c^irder^'Tht IV^'Tr ™-^^^'' ^^ ""- hence they have n'o define , a calcentr ar'"'"""-^' """ .He^~, radical a.s of the^C rrjsTS anXthT:^:, ti:;:et:irntc-s ir :r z no tangent can be drawn from thp r^^ 1 ^ ^^' ^""^ of the circles Tn tht. u ^'^'''^^ ''^"^^^ ^o any one -.n4ti-e::;r::tr:::s^^^^^^^^ tices of a'^Arthr'''"' '"''' """ "^^ "^^^^ f™" 'he ver- A n J ^ ^^^^^^ ^hese lines as R z B diameters. ABC is a A and O its ortho- centre, and AP, BQ, CR are lines trom the vertices to the opposite sides. ^BXA = --|, passes through x; and OX .'o\ tell to ttr " '"""^^^ tangent from O to the circle on Ap' ''' "^"'" "^ ^^^ OF THE RADICAL AXIS. 233 Similarly O Y . OB is the square of the tangent from O to the circle on BQ as diameter, and similarly for OZ.OC But as O IS the polar centre of AABC, (26Q<'^ OX.OA = OY.OB = OZ.OC .-.the tangents from O to the three circles on AP, BQ, and CR are equal, and O is their radical centre. ^.^.,/. Cor. I. Let P, Q, R be collinear. Then the polar centre of AABC is the radical centre of circles on AP, BQ, and CR as diameters. Agam, in the AAQR AP, QB, and RC are lines from the vertices to the opposite sides. .-.the polar centre of AAQR is the radical centre of circles on AP, BQ, and CR as diameters. Similarly the polar centres of the As BPR and CPO are radical centres to the same three circles. But these As have not a common polar centre, as is readily seen. Hence the same three circles have tour different radical centres. And this is possible only when the circles are co-axal. / ..^o ^ . fi, • 1 . (276°, Cor. 5) . . the circles on AP, BQ, and CR are co-axal. .-. if any three collinear points upon the sides of a A be joined with the opposite vertices, the circles on these joins as diameters are co-axal. Cor 2. Since ARPC is a quadrangle or tetragram (247°, Def. 2), and AP, BQ, CR are its three diagonals, .-. the circles on the three diagonals of any quadrangle are co-axal. ° Cor. 3. The middle points of AP, BQ, and CR are col- hnear. But ARPC is a quadrangle of which AP and CR are niternal diagonals, and BQ the external diagonal. .-.the middle points of the diagonals of a complete quad- rangle, or tetragram, are collinear. (See 248°, Ex. 2) Con 4. The four polar centres of the four triangles deter- mined by the sides of a tetragram taken in threes are collinear »(l '' I'ii ft 2.34 SYNTHJiTIC GKOMKTKY. and lie upon the common radical axis of the three circles having the diagonals of the tetragram as diameters. 278°. 77/mr/,/._In general a system of co-axal circles inverts into a co-axal system of the same species. (i.) Let the circles be of the r./.-species The common points become two points' by inversion, and he mverses of all the circles pass through them. Therefore the inverted system is one of ^./.-circles. Cor. I. The axis of the system (LL' of Fig. to 274^) inverts into a circle through the centre of inversion (261°, Cor. i), and as all the inverted circles cut this orthogonally, the axis of the system and the two common points invert into a circle through the centre and a pair of inverse points to it. , (258°, Conv.) Cor. 2. If one of the common points be taken as the centre of inversion, its inverse is at 00 . The axis of the system then inverts into a circle through the centre of inversion, and having the inverse of the other common point as its centre, and all the circles of the system invert into centre-lines to this circle. • (2.) Let the circles be of the /^.-species Let the circles S and S' pass through ihe limiting points and be thus ^./.-circles. ^ Generally S and S' invert into circles which cut the in- verses of all the other circles orthogonally. (26,°) .;. the intersections of the inverses of S and S' are limiting points, and the inverted system is of the /./.-species. Cor. 3. The axis of the system (MM' of Fig. to 274°) be- comes a circle through the centre and passing through the imitmg points of the inverted system, thus becoming one of the f./.-circles of the system. Cor. 4. If one of the limiting points be made the centre of OF THE RADICAL AXIS. 235 inversion, the circles S and S' become centre-lines, and the /./.-circles become concentric circles. Hence concentric circles are co-axal, their radical axis bemgatoo. Exercises. 1. What does the radical axis of (i, 278) become.? 2. What does the radical axis of (2, 278') become.? 3. How would you invert a system of concentric circles into a common system of /^.-circles ? 4. How would you invert a pencil of rays into a system of ^./.-circles. 5- The circles of 277° are common point circles. 279°. rheorem.—kny two circles can be inverted into equal circles. Let S, S' be the circles having radii r and r', and let C, C be the equal circles into which S and S' are to be inverted ; and let the common radius be p. Then PP = /'=OP.OQ OQ r 0(22 • Similarly, P^OP^OQ' ^' r' OQ'2 • But, since P and Q and also P' and Q' are Tnverse points OP.OQ = OP'.OQ', OQ2 r QQ,,=^ = a constant, and (275°, Cor. 5) O lies on a circle co-axal with S and S' And vyith any point on this circle as a centre of inversion S and S mvert into equal circles. Cor. I. Any three non-co-axal circles can be inverted into equal circles. ?8I 1^: 2?>6 SYNTHETIC GEOMETRY. m I o^nL II T^""! ""? ''' ^'' ^'' ^'^"^ ^'' ^ ^^""^^ ^he locus of O for which S and S' invert into equal circles, and Z' the locus of O for wh.ch S and S" invert into equal circles. Then ^ and Z are circles of which Z is co-axal with S and S', and 2 IS co-axal with S and S". And, as S, S', and S";ire not co-axal, Z and Z' intersect in two points, with either of which as centre of inversion the three given circles can be inverted into equal circles. Cor 2. If S, S', and S" be /^-circles, Z and Z' being co-axal with them cannot intersect, and no centre exists with which the three given circles can be inverted into equal circles. But If S, S' and S" be ./.-circles, Z and Z' intersect in the common points, and the given circles invert into centre-lines of the circle of inversion, and having each an infinite radius these circles may be considered as being equal. (278^ Cor. 2) Cor. 3. In: general a circle can readily be found to touch three equal circles. Hence by inverting a system of three circles into equal circles, drawing a circle to touch the three gllen drde"""'"' "' '"'"" ^""^^ "'^^^ ^^^^^ ^^- 280°. Let the circles S and S;, with centres A and B and radii r and r, be cut by the circle Z with centre at O and radius OP = R. Let NLbe the radical axis of S and S'. Since AP is J. to the tangent at P to the circle S, and OP is _L to the tangent the ^P0 = ^ is .he ande o- i---r ^'/ '^ '^ "'''''' ^' -3 -ic cin^ie Ox hucrbection of the circles S OF THE RADICAL AXIS. 237 and Z (115°, Def. i). Similarly BQO = is the angle of intersection of the circles S' and Z. Now PP' = 2rcos^ = R-OP', and QQ' = 2/cos0 = R-OQ', OF-OQ' = 2(r'cos0-rcos^). But R.0P'-R.0Q' = 0'P-0T'2 (where OT is the tangent from O to S, etc.) =2AB . OL, (275") AB R = OL. r'cos^-rcos^ Cor. I. When 6 and are constant, R varies as OL. .'. a variable circle which cuts two circles at constant angles has its radius varying as the distance of its centre from the radical axis of the circles. Cor. 2. Under the conditions of Cor. i ON varies as OL, and .' OL . ON IS constant. .*. a variable circle which cuts two circles at a constant angle cuts their radical axis at a constant angle. Cor. 3. When OL = o, ^cos0 = rcos^, ^^^ r : / = cos : cos 6. .-.a circle with its centre on the radical axis of two other circles cuts them at angles whose cosines are inversely as the radii of the circles. Cor. 4. If circle Z touches S and S', d and are both zero or both equal to tt, or one is zero and the other is t. .-. when Z touches S and S', R-^^^^ . OL, where the variation in sign gives the four possible varieties of contact. Cor. 5. When ^ = 0=|. Z cuts S and S' orthogonally, and OL=o, and the centre of the cutting circle is on the radical axis of the two. Ml :| '1 4% f Ail -f- 2Z^ SVNTHIiTlC GKOMETRV SECTION VII. CENTRE AND AXES OF SIMILITUDE OR PERSPECTIVE. given in Art. 254 . We here propose to extend these reh- tions to the polygon and the circle. 281 B Let O, any point, be connected ^vith the vertices A, "> C, ... of a polygon, and on OA, OB, OC, ... let points a, d, ^, ... be taken so that OA ; Or? = OB : 0^=0C : O^... and OA:0^'=OB:0^'=OC:0^'... Then, since OAB is a A and a/, is so drawn as to divide the sides proportionally in the ?ame order, ••• al? is II to AB. (202°, Conv.) Similarly, ^^islltoBC, rrt'toCD, etc., ^V'islltoBC, ^V to CD, etc.,' AOAB ^AOad^^Oa'd', A0BC^A0^6'^A0^V', . .-.the polygons ABC..., abc..., and a'l^'c'.,. are all similar and have their homologous sides parallel. /^./.-The polygons ABCD... and abed... are said to be similarly placed, and O is their .^/.r;/^/ centre of similitude • whde the polygons ABCD... and a'l/c'd'... are oppositely placed, and O is their hifenial centre of similitude Hence, when the lines joining any point to the vertices of a polygon are all divided in the same manner and in the same order, the points of division are the vertices of a second similarly, and CENTKK OF SIMILITUDE OR I'lL USPKCTIVj:. 239 polygon similar to the original, and so placed that the homologous .,des of the two polygons are parallel. thT'l: ^^'^*'" "™ ''""^'' P°'>S°"^ «« ^o Pl'-'ced as to have he,r homologous sides parallel, they are in perspective and Let ABCD.., akvf... be the polygons Since they are similar, AB : al,= hC : /..= CD : .,/ (.07=^ and by hypothesis AB is || to a^, BC to /,,, etc. '^ ^' Let Aa and Bd meet at some point O. Then OAB is a A and a/; is !| to AB. PB^AB_BC__ , /.^C. passes through O, and similarly D./ passes through O, By writing a'd'c'... for adc... the theorem is proved for the polygon .'^VV'. which is oppositely placed to ABCD.. fir-Zr'J^ ^\^f ^^^' "'''' '"' "^' ^^==^^^' '-^"^^ hence ^^-iiC, etc., and the polygons are congruent. Cor. 2. The joins of any two corresponding vertices as A, ^;^,c; a, c are evidently homologous lines in the polygons and are parallel. ^ ^^ Similarly any line through the centre O, as XxOx' is homologous for the polygons and divides them similarly. 283°. Let the polygon ABCD... have its sides indefinitely fZw, «^? """^'' '"^ diminished in length. Its limiting torn (148 ) ,s some curve upon which its vertices lie A also of T"r" '^' ^'""'"^ '"■" °^ ^'^ P°^>«^""^ -^-••- «s also of « ^ .,/..., smce every corresponding pair of limiting or vanishing elements are similnn Hence if two points en a variable radius vector have the ratio of their distances from the pole constant, the loci of the f; i\ \ r f mi li ■ill I i i i H 240 SYNTHETIC GEOMETRY. points are similar curves in perspective, and having the pole as a centre of perspective or similitude. Cor. I. In the limiting form of the polygons, the line DC becomes a tangent at 13, and the line be becomes a tangent at A And similarly for the line be'. .'. the tangents at homologous points on any two curves in perspective are parallel. 284°. Since a^rrf... and a'b'c'd',.. are both in perspective with ABCD... and similar to it, we see that two similar polygons may be placed in two different relative positions so as to be in perspective, that is, they may be similarly placed or oppo- sitely placed. In a regular polygon of an even number of sides no dis- tinction can be made between these two positions ; or, two similar regular polygons are both similarly and oppositely placed at the same time when so placed as to be in per- spective. Hence two regular polygons of an even number of sides and of the same species, when so placed as to have their sides respectively parallel, have two centres of perspective, one due to the polygons being similarly placed, the external centre ; and the other due to the polygons ^eing oppositely placed' the internal centre. Cor. Since the limiting form of a regular polygon is a circle (148^), two circles are always similarly and oppositely placed at the same time, and accordingly have always two centres of perspective or similitude. 285°. Let S and S' be two circles with centres C, C and radii ;-, r' respectively, and let O and O' be their centres of perspective or similitude. Let a secant line through O cut S in X and Y,and S' in X' and Y'. CENTRE OF SIMILITUDE OR PERSPECTIVE. 241 Then O is the centre of similitude due to considering the circles S and S' as being similarly placed. Hence X and X', as also Y and Y', are homologous points, and (283°, Cor. 1) the tangents at X and X' are parallel. So also the tangents at Y and Y' are parallel. Again O' is the centre of similitude due to con- sidering the circles as being oppositely placed, and for this centre Z and Y' as also U and U' are homologous pomts ; and tangents at Y' and Z are parallel, and so also are tangents at U and U'. Hence YZ is a diameter of the circle S and is parallel to Y'Z' a diameter of the circle S'. Hence to find the centres of similitude of two given circles :— Draw parallel diameters, .one to each circle, and connect their end-points directly and transversely. The direct connector cuts the common centre-line in the external Q I lit Hi Pi iri'r-'-r! 242 SYNTHETIC GEOMETRY centre of similitude, and the transverse connector cuts it in the internal centre of similitude. ! i i 286°. Since OX : OX' = OY : OY', if X and Y become coin- cident, X' and Y' become coincident also. .-. a line through O tangent to one of the circles ir, tangent to the other also, or O is the point where a commor\ tangent cuts tlie common centre-line. A similar remark applies to O'. When the circles exclude one another the centres of similitude are the intersections of common tangents of the same name, direct and transverse. When one circle lies within the other (2nd Fig.) the com- mon tangents are imaginary, although O md O' their points of intersection are real. 287°. Since AOCY^AOC'Y', .-. OC:OC'=r:r', and since AO'CZ « AO'C'Y', .-. O'C : 0'C'=r : r'. .'. the centres of similitude of two circles are the points which divide, externally and internally, the join of the centres of the circles into parts which are .is the conteiminous rndii. The preceding relations give 0C= /' .CC. and 0'C= /' . CC. r-r ' r' + r' : r according as CC is < r' -r, r' + r. r according as CC is OC is and O'C is Hence 1. O lies within the circle S when the distance between the centres is less than the difference of the radii, and O' lies within the circle S when the difference between the centres is less than the sum of the radii. 2. When the circles exclude each other without contact both centres of similitude lie without both circles. 3. When the circles touch externally, the point of contact is the internal centre of similitude. 4. When one circle touches the other internally, the point of contact is the external centre of similitude. CENTRE OF SIMILITUDE OR PERSPECTIVE. 243 5. When the circles are concentric, the centres of similitude coincide with the common centre of the circles, unless the circles are also equal, when one centre of similitude becomes any point whatever. 6. If one of the circles becomes a point, both centres of similitude coincide with the point. 288°. D^'f.—The circle having the centres of similitude of two given circles as end-points of a diameter is called the arc/e of simWhide of the given circles. The contraction ofs. will be used for circle of similitude. Cor. I. Let S, S' be two circles and Z their of ^ Since O and 0' are two points from which tangents to circles S and S' are in the constant / j, ratio of r to r\ the circle o Z is co-axal with S and S' (275", Cor. 5). Hence any two circles and their of s. are co-axal. Cor. 2. From any point P on circle Z, PT :TC = PT':T'C', and.-. z.TPC=^T'PC'. Hence, at any point on the of s. of two circles, the two circles subtend equal angles. Cor. whence OC = CC' 00'=CC', r r' - r 2rr' , and 0'C = CC'. r' + r (287°) •• I. The of s. is a line, the radical axis, when the given circles are equal {r-=r'). 2. The © ofs. becomes a point when one of the two given circles becomes n point (r or r'^o). 3. The ofs. is a point when the given circles are con- centric (CC'=o). i: 11 f -' i •lU its IH 244 SYNTHETIC GEOMETRY. 289°. Def.—Wxih reference to the centre O (Fig. of 285°), X and y, as also X' and Y, are called antihoniologous points. Similarly with respect to the centre O', U' and Z, as also U and Y', are antihomologous points. Let tangents at X and Y' meet at L. Then, since CX is || to C'X', ^CXY = ^C'X'Y' = aC'Y'X'. But ^LXY is comp. of ^CXY and ^LY'X' is comp. of z.C'Y'X'. ALXY' is isosceles, and LX = LY'. L is on the radical axis of S and S'. Similarly it may be proved that pairs of tangents at Y and X', at U and Y', and at U' and Z, meet on the radical axis of S and S', and the tangent at U passes through L. .'. tangents at a pair of antihomologous points meet on the radical axis. Cor. I. The join of the points of contact of two equal tangents to two circles passes through a centre of similitude of the two circles. Cor. 2. When a circle cuts two circles orthogonally, the joins of the points of intersection taken in pairs of one from each circle pass through the centres of similitude of the two circles. 290°. Since OX:OX' = r:r', OX.OY':OX'. OY' = r:/. But OX'. OY' = the square of the tangent from O to the circle S' and is therefore constant. OX . O Y' = -, . or 2 = a constant. r .*. X and Y' are inverse points with respect to a circle whose centre is at O and whose radius is OT' /'',. Def. — This circle is called the circle of antisimilitude^ and will be contracted to of ans. Evidently the circles S and S' are inverse to one another with respect to their of ans. CENTRE OF SIMILITUDE OR PERSPECTIVE. 245 For the centre 0' the product OU . OY' is negative, and the Q)o/ans. corresponding to this centre is imaginary. Cor. I. Denoting the distance CC by d, and the difiference between the radii {r' - r) by 5, we have where R = the radius of the of am. Hence 1. When either circle becomes a point their 0/ ans. becomes a point. 2. When the circles S and S' are equal, the 0/ ans. be- comes the radical axis of the two circles. 3. When one circle touches the other internally the 0/ ans. becomes a point-circle. (^=5.) 4. When one circle includes the other without contact the o/ans. is imaginary. {d<8.) Cor. 2. Two circles and tlieir circle of antisimilitude are co-axal. ^2g^o) Cor. 3. If two circles be inverted with respect to their circle of antisimilitude, they exchange places, and their radi- cal axis being a line circle co-axal with the two circles becomes a circle through O co-axal with the two. The only circle satisfying this condition is the circle of similitude of the two circles. Therefore the radical axis inverts into the circle of similitude, and the circle of simili- tude into the radical axis. Hence every line through O cuts the radical axis and the circle of similitude of two circles at the same angle. .' *... 1!^ 5 ■ ' 291°. Z>^— When a circle touches two others so as to exclude both or to include both, it is said to touch them similarly, or to have contacts of like kind with the two. When it includes the one and excludes the other, it is said to touch them dissiinilarly, or to have contacts of unlike kinds with the two. i\ 246 SYNTHETIC GEOMETRY. 292°. Theorem. — When a circle touches two other circles, its chord of contact passes through their external centre of similitude when the contacts are of like kind, and through their internal centre of similitude when the contacts are of unlike kinds. Proof.— htt circle Z touch circles S and S' at Y and X'. Then CYD and C'X'D are linc3. (113°, Cor. i) Let XYX'Y' be the secant through Y and X'. Then z.CXY = z.CYX=z.DYX' = ^DX'Y = ^CX'Y'. .*. CX and C'X' are parallel, and X'X passes through the external centre of similitude O. (285°) Similarly, if Z' includes both S and S', it may be proved that its chord of contact passes through O. Again, let the circle W, with centre E, touch S' at Y' and S at U so as to include S' and exclude S, and let UY' be the chord of contact. Then ^CVU = _CUV = z.EUY' = ^EY'U, .*. EY' and CV are parallel and VY' connects them trans- versely ; .'. VY' passes through O'. q.e.d. Cor. I. Every circle which touches S and S' similarly is cut orthogonally by the external circle of antisimilitude of S and S'. CENTRE OF SIMILITUDE OR PERSPECTIVE. 24; Cor. 2. If two circles touch S and S' externally their points of contact are concyclic. * (jj^o^ ^.x. 2) But the points of contact of either circle with S and S' are antihomologous points to the centre O. .;. if a circle cuts two others in a pair of antihomologous ponits It cuts them in a second pair of antihomologous points. Cor. 3. If two circles touch two other circles similarly, the radical axis of either pair passes through a centre of simili- tude of the other pair. For, if Z and Z' be two circles touching S and S' externally, the external cn-cle of antisimilitude of S and S' cuts Z and Z' orthogonally (Cor. i) and therefore has its centre on the radical axis of Z and Z'. Cor. 4. If any number of circles touch S and S' similarly they are all cut orthogonally by the external circle of anti- similitude of S and S; and all their chords of contact and all their chords of intersection with one another are concurrent at the external centre of antisimilitude of S and S'. k 293°. 7Vic'on'm.~U the circle Z touches the circles S and S , the chord of contact of Z and the radical axis of S and S' are conjugate lines with respect to the circle Z. Proo/.-Let Z touch S and S' in Y and X' respectively The tangents at Y and X' meet at a point P on the radical axis of S and S'. .gov But P is the pole of the chord of contact YX'. .-. the radical axis passes through the pole of the chord of contact, and reciprocally the chord of contact passes through the pole of the radical axis (267°, Def.) and the lines are conjugate. ¥ 11 li 1248 SYNTHETIC GEOMETRY. AXES OF SIMILITUDE. 294°. Let Sj, S2, S;j denote three circles liavinC, BX ;-., and ri '-.n (; cx CY Ay AZ BZ ra\ CX/ ' and X, Y, Z are col- linear. Similarly it is proved that the triads of points XY'Z', YZ'X', ZX'Y' are collinear. /A/— These lines of collinearity of the centres of similitude of the three circles taken in pairs are the n.vcs of similitude of the circles. The line XYZ is the external axiS; as being external to all the circles, and the other three, passing be- tween the circles, are internal axes. Cor. I. If an axis of similitude touches any one of the circles it touches all three of them. (286°) Cor. 2. If an axis of similitude cuts any one of the circles it cuts all three at the same angle, and the intercepted chords are proportional to the corresponding radii. Cor. 3. Since XYX'Y' is a quadrangle whereof XX', YY', and ZZ' are the three diagonals, the circles on XX', YY', and ZZ' as diameters are co-axal. ('^77°, Cor. 2) .". the circles of similitude of three circles taken in pairs are co-axal. AXES OF SIMILITUDE OR PERSPECTIVE. 249 Cor. 4. Since the three circles of similitude are of the c.p.- species, two points may be found from which any three circles subtend equal angles. These are the common points to the three circles of similitude. (288°, Cor. 2) Cor. 5. The groups of circles on the following triads of segments as diameters are severally co-axal, AX, BY, CZ ; AX, YZ', Y'Z ; BY, Z'X, ZX'; CZ, XY', X'Y. 295°. Any two circles Z and Z', which touch three circles Sj, S2, S3 similarly, cut their circles of antisimilitude ortho- gonally (292", Cor. i), and therefore have their centres at the radical centre of the three circles of antisimilitude. (276°, Cor. 4) But Z and Z' have not necessarily the same centre. .*. the three circles of antisimilitude of the circles S^, S^, and S3 are co-axal, and their common radical axis passes through the centres of Z and Z'. 296°. Theorem. — If two circles touch three circles similarly, the radical axis of the two is an axis of similitude of the three ; and the radical centre of the three is a centre of similitude of the two. Proof.— i:he circles S and S' touch the three circles A, B, and C similarly. I. Since S and S' touch A and B similarly, the radical axis of S and S' passes through a centre of similitude of A and B. (292°, Cor. 3) Also, the radical axis of S and S' passes through a centre of similitude of B and C, and through a centre of similitude of C and A. .'. the radical axis of S and S' is an axis of similitude of the three circles A, B, and C. f-1 !i-f |1 m f; t I ■ n t W' 1) mi 2;o SYNTHETIC GEOMETRY. ! I 2. Again, since A and B touch S and S', the radical axis of A and 13 passes through a centre of similitude of S and S'. For similar reasons, and because A, B, and C touch S and S' similarly, the radical axes of B and C, and of C and A, pass through the same centre of similitude of S and S'. But these three radical axes meet at the radical centre of A, B and C. ' .-. the radical centre of A, B, and C is a centre of simili- tude of S and S'. , ^.t'.a. 297°. Problem.—To construct a circle which shall touch three given circles. In the figure of 296°, let A, B, and C be the three given circles, and let S and S' be two circles which are solutions of the problem. Let L denote one of the axes of similitude of A, B, and C, and let O be their radical centre. These are given when the circles A, B, and C are given. Now L is the radical axis of S and S' (296°, i), and O is one of their centres of similitude. But as A touches S and S' the chord of contact of A passes through the pole of L with respect to A (293°). Similarly the chords of contact of B and C pass through the poles of L with respect to B and C respectively. And these chords are concurrent at O. ("202°^ Hence the following construction :— Find O the radical centre and L an axis of similitude of A, B, and C. Take the poles of L with respect to each of these circles, and let them be the points A g, r respectively. Then op, Og, Or are the chords of contact for the three given circles, and three points being thus found for each of two touching circles, S and S', these circles are determined. (This elegant solution of a famous problem is due to M. (jcrgonne.) Cor. As each axis of similitude gives different poles with respect io A, B, and C, while there is but one radical centre AXES OF SIMILITUDE OR PEKSl'ECTIVii. 25 I O, in general each axis of similitude determines two touching circles ; and as there are four axes of similitude there are eight circles, in pairs of twos, which touch three given circles. Putting / and c for internal and external contact with the touching circle, we may classify the eight circles as follows : (See 294°) Axes OF Similituui£. A B X Y Z • t e e X Y' Z' e i i e X' Y Z' • t e e i X' Y' Z • t e I e :] I pr. c e\ i\ e) % .J4 pr. •2 pr. •3 pr. u. m \\\'k '^': J 511 : PART V. ON HARMONIC AND ANHARMONIC kATlOS— HOMOGRAPHY. INVOLUTION, ETC. SECTION I. GENERAL CONSIDERATIONS IN REGARD TO HARMONIC AND ANHARMONIC DIVISION. 298°. Let C be a point dividing a segment AB. The posi- tion of C in relation to A and B is determined by the ratio I I I . AC : BC. For, if we know this ratio, we ^ ^ ^ know completely the position of C with respect to A and B. If this ratio is negative, C lies between A and B ; if positive, C does not lie between A and B. If AC:BC=-i, C is the internal bisector of AB ; and if AC : BC= + I, C is the external bisector of AB, /.c, a point at 00 in the direction AB or BA. Let D be a second point dividing AB. The position of D is known when the ratio AD : BD is known. De/.—U we denote, the ratio AC : BC by ;;/, and the ratio AD : BD by ;/, the two ratios m : ;/ and n : m, which are reciprocals of one another, are called the two anharmonic ratios of the division of the segment AB by the points C and D, or the harmonoids of the range A, B, C, D. 252 s e HARMONIC AND ANHARMONIC DIVISION. 253 Either of the two anharmonic ratios expresses a relation between the parts into which the segment AB is divided by the points C and D. Evidently the two anharmonic ratios have the same sign, and when one of them is zero the other is infinite, and vice versa. These ratios may be written : — I. 2. ^•^ or ^^-^ P nr AC.BD liC -BD BC.AD ^^ ADTbC' ^P • AC ^^ AD . BC ^^ AD . BC BD'BC BD.AC°' AC.BU' The last form is to be preferred, other things being convenient, on account of its symmetry with respect to A and B, the end-points of the divided segment. i \4 '■*.,. ! ' V 299°. The following results readily follow. '• ^"^ ad". BC ^^ +• '^^'" Fc '''"^ IE ^^""^ ^^^ ''-"^' and therefore C and D both divide AB internally or both externally. ^2^go^ In this case the order of the points must be some one of the following set, where AB is the segment divided, and the letters C and D are considered as being interchangeable : CDAB, ACDB, CABD, ABCD. _ T^, AC.BD, ^, AC , AD, ^' ^^^ AD . BC ^^ "• ^^^'' ^ ''^"^ Si; ^^' ' °PP«s5te signs, and one point divides AB internally and the other externally. Tb'^ order of the points is then one of the set CADB, ACBD. 3. When either of the tvvu anharmonic ratios is ±1, these ratios are equal. AC.BD . . ^, AC AD 4. Let AD r)V- .= + !. Then r^ = . --. an BC BD d C and D are both internal or both external. \\% ■ f 254 Also SYNTIIKTIC GFOMKTRY. AC-HC AD-HD ATi AB - , or - , T . AC.BI) 5-^^^^AD.BC=-'- Then BC Jil) '^' HC Miy and C and D coincide. Hence, when C and I) are distinct points, the anharmonic ratio of the parts into which C and I) divide Ali cannot be positive unity. AC^AD lic my And since C and D are now one external and one internal (2), they divide the segment AH in the same ratio internally and externally, disregarding sign. Such division of a line segment is called harmonic. (208°, Cor. i) Harmonic division and harmonic ratio have been long em- ployed, and from being only a special case of the more general ratio, this latter was named "anharmonic" by Chaslcs, "who was the first to perceive its utility and to apply it extensively in Geometry." 300°. /;^ \ \\ "% V 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 '^'l-- ^ 268 SYNTHETIC GEOMETRY. 0{ECQF} = C{EOQF} -C{RDBF} = A{RDBF} = A{EDPF} = 0{EDPF} = 0{ECPF}. (referred to axis EF) (referred to axis DR) (referred to axis DE) (by reversing rays, etc.) .'. the pencils O. ECQF and O. ECPF are equianharmonic, and having three rays in common the fourth rays must be in common, z>., they can differ only by a straight angle, and therefore O, P, Q are collinear. (Being the first application of anharmonic ratios the work is very much expanded.) .'.If six lines taken in order intersect alternately in two sets of three collinear points^ they intersect in a third set of three collinear points. Cor. I. ABC and DEF are two triangles, whereof each has one vertex lying upon a side of the other. If AB and DE are taken as corresponding sides, A and F are non-corresponding vertices. But, if AB and EF are taken as corresponding sides, A and D are non-corresponding vertices. Hence the intersections of AB and EF, of ED and CE, and of AD and CF are collinear. .'. If two triangles have each a vertex lying upon a side of the other, the remaining sides and the joins of the remaining non-corresponding vertices intersect collinearly. Cor. 2. Joining AD, BE, CF, ADBE, EBFC, and ADFC are quadrangles, and P, Q, O are respectively the points of intersection of their internal diagonals. .*. if a quadrangle be divided into two quadrangles, the points of intersection of the internal diagonals of the three quadrangles are collinear. OF ANHARMONIC PROPERTIES. i6g 316°. Let A, A', B, B', C, C be six points lying two by two on two sets of three con- current lines, which meet at P and Q. Then the points lie upon a third set of three concurrent lines meeting at O. We are to prove that AO and AA' are in line. A{0;rC'B}=B'{OK:'Q} = B'{CPC>} = Q{A'P^B} =A{A'jrC'B}. .-. the pencils A.O;rC'B and A.A';irC'B are equi- anharmonic, and have three corresponding rays in common. Therefore AO and AA' are in line. Cor. ABC and A'B'C are two As which are in perspective at both P and Q, and we have shown that they are in per- spective at O also. As there is an axis of perspective corresponding to each centre, the joins of the six points, accented letters being taken together and unaccented, together, taken in every order intersect in three sets of three collinear points. Exercises. 1. If two As have their sides intersecting collinearly, their corresponding vertices connect concurrently. 2. The converse of Ex. i. 3. Three equianharmonic ranges ABCD, A'B'C'D', and PQRS have their axes concurrent at Y, and AA', BB', v^v- , j^i^ i,vjin-UiiciiL '} = 0{ABCD}, {abcd}^{^BCT>}, AC.JBD^ae^bd AB. CD ab.ed' But AC . BD =the square on the common direct tangent to S and S', and ae. bd=ihe square of the corresponding tangent to J and/. (179°, Ex. 2) And AB . CD and ab.cdave the products of the diameters respectively. or II III! fore the es of an h the X ratios of of the on their hanged, be the ) be the re hne, ircles s e circle inverses s and s' md B'C'D'}. be the 2 line of dB\ cC\ ncurrent ', Ex. i) jent to S ■ tangent )% Ex. 2) iameters OF ANHARMONIC PROPERTIES. 273 And the theorem is proved. Cor. I. Writing the symbolic expressions {ABCDJ and {abed] m another form, we have AB.CD ab.cd' And AD . EC and ad. be are equal to the squares on the transverse common tangents respectively,, Cor. 2. If four circles S„ S^, S3, S4 touch a line at the ponns A, B, C, D, and the system be inverted, we have four circles Ji, ^2, j-3, s^, which touch a circle Z through the centre of mversion. Novv let d,, d, ^3, d, be the diameters of S„ S^, etc., and let ^1, S h. h be the diameters of Sy, s„ etc., and let /,., be the common tangent to s, and s,, t,, be that to ^3 and s,, ac Then AB, etc., are common tangents to Sj and S^, etc.. and And AB.CD + BC.AD + CA.BD=o. ABa^V (^~33l CD2_/3,2 d^d^ 8^8^ sJd^i^d^d^ sJW^^8l and similar equalities for the remaining terms, ^12^34 + ^23^14 + ^31^24 = 0. This theorem, which is due to Dr. Casey, is an extension of Ptolemy's theorem. For, if the circles become point- circles, the points form the vertices of a concyclic quadran-le and the tangents form its sides and diagonals. " If we take the incircle and the three excircles of a trianoie as the four circles, and the sides of the triangle as tangems we obtain by the help of Ex. i, 135°, 32. ^2+,-2_^,2+^2_/,2 as the equivalent for /,2^31 + etc.; and as this expression is identically zero, the four circles given can all be touched bv a fifth circle. S ■w% 11 i Mi :( ;; i \A i i 319° SYNTHETIC Gi:OMETRY. * Let A, 15, C, D, K, F be six points on a circle so con- nected as to form a hexagram, i.e., such that each point is con- nected with two others. Let the opposite sides AB, dp: meet in P ; BC, EF in Q ; and CD, FA in K. To prove that P, Q, and R are collinear. ()!HDER} = Q{CDER} = FiCDEA} = B1CDEPI = Q!BDEP1, .-. Q Rand QP are inline. .-. if a hexagram have its vertices concyclic, the points of intersection of its opposite sides jp in pairs are collinear. /;,,/; _The line of collinearity is called the pascal of the hexagram, after the famous Pascal who discovered the theorem, and the theorem itself is known as Pascal's theorem. Cor. I. The six points may be connected in 5 x 4 x 3 x 2 or 120 different ways. For, starting at A, we have five choices for our first connection. It having been fixed upon, we have four for the next, and so on to the last. But one-half of the hexagrams so described will be the other half described by going around the figure in an opposite direction. Hence, six points on a circle can be connected so as to form 60 different hexagrams. Each of these has its own pascal, and there are thus 60 pascal lines in all. When the connections are made in consecutive order about the circle the pascal of the hexagram so formed falls without the circle ; uut if any other order of connection is taken, the pascal may cut the circle. ^=f: OF AXIIARMONIC PROPKRTI KS. 275 Cor 2. In the hexagram in the figure, the pascal is the line through P, O, R cutting the circle in H and K. Now cjkfbd;-c;kfori e = FIKCOR| = F!KCEA}, .-. ^ IKFBD| = |KCEA}, and K is a common point to two equi- "inV p/\ /r\/\q /k anharmonic systems on the circle. So also is H. These points are important in the theory of homographic systems. ^. k ^ Cor. 3. Let i, 2, 3, 4, 5, 6 denote six points taken consecu- tively upon a circle. Then any particular hexagram is denoted by wntmg the order in which the points are connected, as for example, 2461352. In the hexagram 246135 the pairs of opposite sides are -4 and 13, 46 and 35, 61 and 52, and the pascal passes through then- mtersections. Now taking the four hexagrams 246135, 245136, 246315, 245316, the pascal of each passes through the intersection of the connector of 2 and 4 with the connector of i and 3. Hence the pascals of these four hexagrams have a commr n point. It is readily seen that inverting the order of 2 and 4 gives hexagrams which are only those already written taken in an inverted order. .-. the pascals exist in concurrent groups of four, meeting at fifteen points which are intersections of connectors. Cor. 4. In the hexagram 1352461 consider the two triangles forme^by the side^ 13^52, 46 and 35, 24, 61. The sides 13 and 24, 35 and 46, 52 and 61 intersect on the pascal of 1352461, and therefore intersect collinearly. _ Hence the vertices of these triangles connect concurrently, ^.e., the Ime through the intersection of 3s and 6? and the intersection of 52 and 46, the line through the intersection of i*< . i m SYNTHETIC GEOMETRY. 35 and '24 and the intersection of [3 and 46, and the line through the intersections of 24 and 61 and the intersection of 13 and 25 are concurrent. But the fust of these lines is the pascal of the hexagram 1643521, the second is the pascal of the hexagram 3564213* and the third is the pascal of the hexagram 4256134. .-. the pascals exist in concurrent groups of three, meeting at 20 points distinct from the 1 5 pomts already mentioned. Cor. 5. If two vertices of the hexagram coincide, the figure be- comes a pentagram, and the missing side becomes a tangent. .-. if a pentagram be inscribed in a circle and a tangent at any vertex meet the opposite side, the point of intersection and the points where the sides about that vertex mee^ the remaining sides are coUinear. Ex. I. The tangents at opposite vertices of a concycHc quad- rangle intersect upon the external diagonal of the quadrangle. Ex. 2. ABCD is a concyclic quadrangle. AB and CD meet at E, the tangent at A meets BC at G, and the tangent at B meets AD at F. Then E, F, G are collinear. 320°. Let six tangents denoted by the numbers i, 2, 3, 4, 5, and 6 touch a circle in A, B, C, D, E, and F. And let the points of inter- section of the tangents be de- noted by 12, 23, 34, etc. Then the tan- gents form a hexagram about the circle. Now, 12 is the pole of AB, and 45 is the pole of ED. Therefore the OF ANIIARMONIC PROPERTIES. he line rsection xagram 564213* meeting oned. gure be- :angent. igent at rsection lee*^ the ic quad- drangle. md CD tangent , 3, 4, 5» touch a A, B, C, and F. et the )f inter- of the ; be de- y 12, 23, the tan- form a m about e. 12 is ifore the ^77 line 12.45 is the polar of the point of intersection of AB and ED. Similarly the line 23.56 is the polar of the intersection of BC and EF, and the line 34.61 is the polar of the intersec- tion of CD and FA. But since ABCDEF is a hexagram in the circle, these three intersections are coUinear. (319°) .-. the lines 12 . 45, 23 . 56, and 34. 61 are concurrent at O. And hence the hexagram formed by any six tangents to a circle has its opposite vertices connecting concurrently. /?'C'1)';. Similarly, {ABCEJ = JA'B'C'E'I, {BCDEJ = JB'C'D'E'}, etc. HOMOGRAPHV AND INVOLUTION. 285 Evidently for each position of P, P' can have only one position, and conversely, and hence the points of division on the two axes correspond in unique pairs. .-. two lines are divided homoj,naphically by two sets of points when to each point on one corresponds one and only one point on the other, and when any four points on one line and their four correspondents on the other form equianhar- moniq ranges. Cor. I. If the systems of points be joined to any vertices O and O', the pencils O.ABCD... and 0'. A'B'C'D'... are evidently homographic, and cut all transversals in homo- graphic ranges. Cor. 2. The results of Arts. 304°, Cors. 3 and 4, and of Arts. 305° and 306" and their corollaries are readily extended to homographic ranges and pencils. ,'b I fit The following examples of homographic division are given. Ex. I. A line rotating about a fixed point in it cuts any two lines homographically. Ex. 2. A variable point confined to a given line determines two homographic pencils at any two fixed points. Ex. 3. A system of r._^.-circles determines two homographic ranges upon any line cutting the system. Consider any two of the circles, let P, Q be the common points, and let the line L cut one of the circles in A and A' and the other in B and B'. Then the z.PBB' = z.PQB', and _PAB'=^PQA'. .•.^APB=^A'(2B'. Hence the segment 13A subtends the same angle at P as the segment at B'A' does at O. And similarly for all the segments made in the other circles. Ex. 4. A system of /./^.-circles determines two homographic ranges upon every line cutting the system. <'i 284 SYNTHETIC GEOMETRY. DOUBLE POINTS OF HOMOGRAPHIC SYSTEMS. ill 328'. Let ABCD. A . B and A'B'C'D'... be two homographic _^____^ c p ranges on a common axis. ° A' B' c D' If any two correspondents from the two ranges become coincident the point of co- incidence is a double point of the system. If A and A' were thus coincident we would have the rela- tions {ABCE}-{AB'C'E'}, etc. Thus a double point is a common constituent of two equi- anharmonic ranges, of which the remaining constituents are correspondents from two homographic systems upon a common axis. ABC and A'B'C being fixed, let D and D' be two variable correspondents of the doubly homographic system. Then {ABCD}={A'B'C'D'}, whence BD . A^D^ _ BC^A/C' _^ say. AU.B'D' AC.B'C q' Now taking O, an arbitrary point on the axis, let OD-;i-, OD'=y, OK=a, OM=a\ 0B = <5, 0B'=^'. ThenBD=.r-/;, B'B'=x'-d\ AD=.r-rt, M\^'=x'^a', (x-l>) { x'-a') _p {x-a){x'-b') q' which reduces to the form .m'' + P;r-f-Q;ir' + R=0. When D and D' become coincident x' becomes equal to x and we have a quadratic from which to determine ;r, i.e., the positions of D and D' when uniting to form a double point. Hence every doubly homographic system has two double points which are both real or both imaginary, and of which both may be finite, or one or both may be at infinity. Evidently there cannot be more than two double points, for since such points belong to two systems, three double points would require the coincidence of three pairs of corre- spondents, and hence of all. (306°) IIOMOGRAPHY AND INVOLUTION. 285 329°. If D be one of the double points of a doubly homo- graphic system, J^^-^A-'=CB . CA'^ P Now DB.DA' and DA.DB' are respectively equal to the squares on tangents from D to any circles passing through B, A' and B', A. But the locus of a point from which tangents to two given circles are in a constant ratio is a circle co-axal with both. (275°, Cor. 5> Hence the followmg construction for finding the double points. Through A, B' and A', B draw any two circles so as to mtersect in two points U and V, and through these points of intersection pass the circle S", so as to be the locus of a point from which tangents to the circles S and S' are in the given ratio y/P : ,/Q. The circle S" cuts the axis in D, D, which are the required double points. Evidently, instead of A, B' and A', B we may take any pairs of non-corresponding points, as A, C and A', C; or B, C and B', C. The given ratio ^F : ^Q is different, however, for each different grouping of the points. Cor. I. When P = Q, i.e., when |^ = g!§l the circle S" takes its limiting form of a line and cuts the axis at one finite- point or at none. In this case both double points may be at 00 or only one of them. Cor. 2. If any disposition of the constituents of the system causes the circle S" to lie whoiiy upon one side of the axis, the double points for that ''■^position become imagin- ary. ^iP 4*1 illfl '11 % 2S6 SYNTH r':TIC (JEOMETKY. ^1 330°. Let L be the axis of a doubly homographic system. , Throuj^h any point O on the circle S transfer the system, by rectilinear projection, to the circle. Tiien ABC..., ^ A'B'C... form a doubly homographic system on the '^' circle. D Now, by connecting any two pairs of non-correspond- B ents A, B' and A', ]} ; ]}, C' and B', C ; C, A' and C, A, we obtain the pnscal line KH which cuts the circle in two points such that {KABC} = {KA'B'C'J. Hence H and K are double points to the sys^tlm or'the circle. And by transferring K and H back through the point O to the axis L, we obtain the double points D, D of the doubly homographic range. Cor. I. When the pascal falls without the circle, the double points are imaginary. Cor. 2. When one of the joins, KO or HO, is || to L, one of the double points is at 00 . Cor. 3. If the system upon the circle with its double points H and K be projected rectilinearly through any point on the circle upon any axis M, it is evident that the projected system is a doubly homographic one with its double points. Cor. 4. Cor. 3 suggests a convenient method of finding the ; B, C d C, A, cal line circle in irc'}. Cor. 2) on the le point of the ! double , one of ; points )int on ejected oints. ing the employ ny pair r I HOMr)(;KAI>IIY AND INVOLUTION. 287 Then from any convenient point on the circle transfer the rema.nmg points, lind the pascal, and proceed as before. 331". The following are examples of the application of the double pomts of doubly homographic systems to the solution of problems. K.x. I. (;ivcn two non-parallel lines, a point, and a third hue To place between the non-parallels a segment which sha I subtend a given angle at the given point, and be parallel to the third line. Let L and M be the non- parallel lines, and let N be the third line, and () be the given point. We are to place a segment between L and M, so as to subtend a given angle at O and be |) to N. On L take any three points A, B, C, and join OA, OB OC Draw Aa, B/., CV all || to N, and draw Oa\ O//, O.-' so as to make the angles AO..', liO^', CO.' each equal to the given angle. ^ Now, if with this construction a coincided with a', or l> with //, or c with c', the problem would be solved. But, if we take a fourth point D, we have 0{ABCD} == lABCD] = {abed} = {a'b'dd']. .'. ridc'd and a7/c'd' nve two homographic systems upon the same axis. Hence the double points of the system give the solutions required. Ex. 2. Within a given A to inscribe a A whose sides shall be parallel to three given lines. Ex. 3. Within a given A to inscribe a A whose sides may pass through three given points. Ex. 4. To describe a A such that its sides shall pass through three given points and its vertices lie upon three given lines. H' k I: ii.t i f Hi ni 288 SYNTHETIC GEOMETRY. SYSTEMS L\ INVOLUTION. 332°. If A, A', B, B' are four points on a common axis, whereof A and A', as also B and B', are correspondents, a pomt O can always be found upon the axis such that OA.OA' = OB.OB'. This point O is evidently the centre of the circle to which A and A', and also B and B', are pairs of inverse points, and is consequently found by 257°. Now, let P, P' be a pair of variable conjugate points which so move as to preserve the relation OP.OP' = OA.OA' = OB.OB'. Then P and P' by their varying positions on the axis deter- mine a double system of points C, C, D, D', E, E', etc. conjugates in pairs, so that OA . OA'=OC. OC=^OD . OD' = OE . OE' = etc. Such a system of points is said to be in involution, and O is called the centre of the involution. When both constituents of any one conjugate pair lie upon the same side of the centre, the two constituents of every conjugate pair lie upon the same side of the centre, since the product must have the same sign in every case. With such a disposition of the points the circle to which conjugates are inverse points is real and cuts the axis in two '^. -rr—^ — III — . , , , points F and F'. C FCOABFB' A. , At these points vari- able conjugates meet and become coincident. Hence the points F, F' are the double points or ^^/ of the system. From Art. 311°, r, FF' is divided harmonically by every pan- of conjugate points, so that FAF'A', FBF'B', etc., are all harmonic ranges. When the constituents of any pair of conjugate points lie upon opposite sides of the centre, the foci are imaginary. HOMOGRAPHY AND INVOLUTION. 289 I .^rfu ^u ^' ^'' ^' ^''' ^' ^' ''^ '''' P°"^^s in involution, and let U be the centre. Draw any line OPQ through U, and take P and O so that 0Pr'0Q = 0A.0A', and join PA, PB, PC, and PC, and also QA', QB', QC, and QC. Then, •,• OA. OA' = OP. OQ, o .•. A, P, Q, A' are concyclic. .'. ^OPA = ^OA'Q. Similarly, B, P, Q, B' are concyclic, and Z.OPB=z.OB'o' etc z.APB=/.A'QB'. Similarly, ^BPC = ^B'(2C', ^CPC' = z.C'QC, etc Hence the pencils P(ABCC') and Q(A'B'C'C) are equianhar- monic, or {ABCC} = {A'B'CCf, Hence also {ABB'C} = {A'B'BCT}, {AA'BC} = {A'AB'C}. And any one of these relations expresses the condition that the SIX pomts symbolized may be in involution. 334°. As involution is only a species of homographv, the relations constantly existing between homographic ranges and their corresponding pencils, hold also for ranges and pencils m involution. Hence 1. Every range in involution determines a pencil in involu- tion at every vertex, and conversely. 2. If a range in involution be projected rectilinearly through any point on a circle it determines a system in involution on the circle, and conversely. Ex. The three pairs of opposite connectors of any four points cut any line in a six-point involution. A, B, C, D are the four points, s j d c and P, F the line cut by the six connectors CD, DA, AC, CB, BD, and AB. Then D{PQRR';=D{CARB} b -BfCARD} i ;)i{ ; h* (. ■vti = B{Q'P'RR'} = {P'Q'R'R|^ (302°) if;. 290 SYNTHETIC OLOMKTRY. {P0RR'} = {1>'()'R'R}, and the six points arc in involution. <. Cor. I. The centre O of the involution is the radical centre of any three circles through PP', (^O', and RR'; and the three circles on the three segments PP', OQ', and RR' as diameters are co-axal. When the order of POR is opposite that of P'Q'R' as in the rigure, and the centre O lies outside the points, the co-axal circles are of the /./.-species, and when the two triads of ponits have the same order, the co-axal circles are of the f./.-species. Cor. 2. Considering ABC as a triangle and AD, BD, CD three lines through its vertices at D, we have— The three sides of any triangle and three concurrent lines through the vertices cut any transversal in a six-point involution. Exercises. 1. A circle and an inscribed quadrangle cut any line through them in involution. 2. The circles of a co-axal system cut any line through them in involution. 3. Any three concurrent chords intersect the circle in six points forming a system in involution. 4. The circles of a co-axal system cut any other circle in involution. 5. Any four circles through a common point have their six radical axes forming a pencil in involution. lical centre '; and the (\d RR' as V as in the he co-axal ) triads of are of the , BD, CD rrent lines six-point B through ugh them :le in six circle in their six INDEX OF DEFINITIONS, TERMS, ETC T/ie Nmnbcrs refer to the Articles, Addition Theorem for Sine and Cosine, . Altitude, . Ambiguous Case, Angle, . ,, Acute, . ,, Adjacent internal ,, Basal, . ,, External, ,, Obtuse, ,, Re-entrant, Right, . Straight, Arms of, Bisectors of, Complement of, Cosine of. Measure of, Sine of. Supplement of. Tangent of, \''ertex of, Angles, Adjacent, »» Alternate, M Tnteradjacenl, »» »» Opposite, 236 87 66 31 40 49 49 49 40 89 36 36 32 43 40 213 207 213 40 213 32 35 11 73 39 Angles, \'crticjil, . . 3^ , , Sum and Difference of, 35 ,, of the same Affection, 65 Anharmonic Ratio, . Antihomologous, Apothem, Arc, Area, f> of a Triangle,. Axiom, . Axis of a Range, M Perspective, . .» Similitude, . .. Symmetry, . Basal Angles, . Biliteral Notation, . Bisectors of an Angle, Brianchon's Theorem, Brianchon Point, Centre of a Circle, . t. Inversion, M Mean Position, 1 erspective, Similitude, 298 289 146 lOI 136 •75i 3 230 254 294 lOI 49 22 43 320 320 92 256 238 254 281 ii: 291 292 SYNTHETIC (iKOMETRY. Centre-line, Centre-locus, . Centroid, Circle, Circle of Antisiniilitiule, ,, Inversion, . M Similitiule, Circumanyle, . Circumcentre, . Circumcircle, . Circumference, Circumradius, . Circumscribed Figure, Chord, . Chord of Contact, . Co-axal Circles, CoUinear Points,: . i: Commensurable, Complement of an Angle Concentric Circles, . Conclusion, Concurrent Lines, . I Congruent, Concyclic, Conjugate Points, etc.. Contact of Circles, . Continuity, Principle of, Corollary, C.-P. Circles, . Cosine of an Angle, Constructive Geometry, Curve, Datum Line, ... 20 Degree 41 Desargue's Theorem, . 253 Diagonal, . . 80, 247 Diagonal Scale, . . jiii 95 129 85 92 290 256 288 36 86,97 97 92 97 97 95 114 273 , 247 150 40 93 4 5. 247 51 97 267 291 104 8 274 213 117 15 Diameter, ... 95 Difference of Segments, . 29 Dimension, ... 27 Double Point, . . 109, 328 Eidograph, End-points, Envelope, Equal, . . . : Equilateral Triangle, Excircle, . External Angle, Extreme and Mean Ratio Extremes, Finite Line, Finite Point, . Generating Point, . Geometric Mean, Given Point and Line, Harmonic Division, Harmonic Ratio, Harmonic Systems, . Homogeneity, . nomographic Systems, Homologous Sides, . Homologous Lines and Points, . . .196 Hypothenuse, . . 88, 168 Hypothesis, ... 4 Incircle of a Triangle, . 131 Incommensurable, . .150. Infinity, . . .21, 220 Initial Line, ... 20 Inscribed Figure, . . 97 211.^ 22 223 7, 136 53 131 49 183 193 21 21 69 169 20 208 299 3, 314 160 327 196 m iii' • 95 , . 29 109, 328 21 1 i 22 223 7, 136 53 131 49 183 193 tio ind 88 21 21 69 169 20 208 299 3. 314 160 327 196 196 168 131 150 , 220 20 97 INDEX OF DEFIN Interadjacent Angles, . 73 Inverse Points, . 179, 256 260 332 53 ITIONS, TERMS, ETC 293 I Ortliogonal rrojection, 167, 229 Inverse Figures, Involution, Isosceles Triangle, Limit, Limiting Points, Line, Line in Opposite Senses, i ine-segment, . Locus, L.-P. Circles, . Magnitude, Major and Minor, Maximum, Mean Centre, . Mean Proportional, Means, . Measure, Median, . Median Section, Metrical Geometry, Minimum, 167 148 274 12 156 21 69 274 190 102 175 238 169 193 150 55 183 150 175 Nine-points Circle, rf' ^''- ^ 1265, Ex. 2 Normal Ouadrangle, . 89 Obtuse Angle, . Opposite Angles, . Opposite Internal Angles, Origin, Orthocentre, Orthogonally, 40 39 49 20 S8 115 Pantagraph, . Parallel Lines, Parallelogram, Pascal's Hexagram, Pascal Line, . Peaucellier's Cell, Pencil, . Perigon, . Perimeter, Perspective, Perspective, Axis of, ,, Centre of. Perpendicular, . Physical Line, . Plane, Plane Geometric Figure, Plane Geometry, Point, Point of Bisection, ,, Contact, Point, Double, Pole, . Polar, Polar Reciprocal, Polar Circle, Centre, etc.. Polygon, . Prime Vector, . Projection, Proportion, Proportional Compasses, Protractor, 211^ 70 80 319 319 211 203 36 146 254 254 254 40 12 10, 17 10 II 13 30 109 109, 328 32, 266 . 266 . 268 266 132 20 167 192 2Ili 123 Quadrangle, -j Quadrilateral, j ' ^°' ^9 Quadrilateral, complete, . 247 It I M; i II 1 I' f f| \ i r hi.. 294 SYNTHETIC GEOMETRY. ■Hi Radian, . . . Radical Axis, . Radical Centre. Radius, . Radius Vector, Range, . Ratio, Reciprocation, Reciprocal Segments, Rectangle, Rectilinear Figure, . Reductio ad absiirdum. Re-entrant Angle, . Rhombus, Right Angle, . Right Bisector, Rotation of a Line, Rule, . . Rule of Identity, Scalene Triangle, . Secant, . Sector, . Segment, Semicircle, Self-conjugate, a Self-reciprocal,/ Sense of a Line, Sense of a Rectangle, Similar Figures, Similar Triangles, . Sine of an Angle, . Spatial Figure, Square, . . 207 178, 273 276 92 32 230 188 321 184 82 14 54 89 82 36 42 222 16 7 Straight Angle, „ Line, . M Edge, Sum of Line-segments, Sum of Angles, Superposition, . 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