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(T\ U "7 '^- 0-.H ^ Ti . . V ■; 4 C. t a, ^ ^ C( 3S00 ««< Ay h { PKEFACE. To preserve Euclid's order, to supply omissions, to remove defects, to give brief notes of explanation and simpler methods of proof in cases of acknowledged difficulty — such are the main objects of this Edition of the Elements. The work is based on the Greek text, as it is given in the Editions of August and Peyrard. To the suggestions of the late Professor De Morgan, published in the Companion to the British Almanack for 1849, I have paid constant deference. A limited use of symbolic representation, wherein the symbols stand for words and not for operations, is generally regarded as desirable, and I have been assured, by the highest authorities on this point, that the symbols employed in this book are admissible in the Examinations at Oxford and Cambridge.* I have generally followed Euclid's method of proof, but not to the exclusion of other methods recom- * I regard this point as completely settJod iu Cambridge by the following notices prefixed to the i»apers on Euclid set in the Senate-House Examinations : I. In the Previous Examination : Tn anmoerft to these questions any intelligible symbols and abbre- viations may be used. II. In the Mathematical Tripos r In answers to the questions on Euclid the symbol — must not be used. The only abbreviation admitted for the square on AB i$ *'sq. on AB," and for the rectangle contained by AB and CD, **rect. AB, CD." // 1/ viu PREFACE. mended by their simplicity, such as the demonstrations by which I propose to replace (at least for a first read- ing) the difficult Tlieorems 5 and 7 in the First Book. I have also attempted to render many of the proofs, as for instance Propositions 2, 13, and 35 in Book I., and Proposition 13 in Book II., less confusing to the learner. In Propositions 4, 5, 6, 7, and 8 of the Second Book I have ventured to make an important change in Euclid's mode of exposition, by omitting the diagonals from the diagrams and the gnomons from the text. In the Third Book I have deviated with even greater boldness from the precise line of Euclid's method. For it is in treating of the properties of the circle that the importance of certain matters, to which reference is made in the Notes of the present volume, is fully brought out. 1 allude especially to the appli- catioi) of Superposition as a test of equality, to the conception of an Angle as a magnitude capable of unlimited increase, and to the development of the methods connected with Loci and Symmetry. The Exercises have been selected Avith considerable care, chiefly from the Senate House Examination Papers. They are intended to be progressive and easy, 80 that a learner may from the first be induced to work out something for himself. I desire to express my thanks to the friends who have improved this work by their suggestions, and to beg for further help of the same kind. J. HAMBLIN SMITH Cambridob, 1878L ELEMENTS OF GEOMETRY. and when INTRODUCTORY REMARKS. When a block of stone is hewn from the rock, we call it a Solid Body. The stone-cutter shapes it, and brings it into that which we call regularity of form ; and then it becomes a Solid Figure. Now suppose the figure to be such that the block has six flat sides, each the exact counter- part of the others ; so that, to one who stands facing a corner of the block, the three sides which are visible present the appearance re- presented in this diagram. Each side of the figure is called a Svrfar smoothed and polished, it is called a Plane Surt'a. •. The sharp and well-defined edges, in which each pair of sides meets, are called Lines. The place, at which any three of the edges meet, is called a Point A Magnitude is anything which is made up of parts in any way like itself. Thus, a line is a magnitude ; because we mny regard it as made up of parts which are themselves lines. The properties Length, Breadth (or Width), and Thickness ''or Depth or Height) of a body are called its Dimensions. "We make the following distinction between Solids, Surfaces Lines, and Points : A Solid has three dimensions, Length, Breadth, Thickness. A Surface has two dimensions, Length, Breadth. A Line has one dimension. Length. A point Jias no dimensions. s. B. BOOK I. DEFINITIONS. I. A Point is that which has no parts. This is equivalent to saying that a Point has no magnitude, itince we define it as that which cannot be di vided into smaller parts. II. A Lime is length without breadth. We cannot conceive a visible line without breadth ; but we can reason about lines as if they hud no breadth, and this is what Euclid requires us to do. III. The Extremities of finite Lines are points. A point marks positiorif as for instance, the place where %, line begins or ends, or meets or crosses another .line. IV. A Straight Line is one which lies in the same direction from point to point throughout its length. V. A Surface is that which has length and breadth only. VI. The Extremities of a Surface are lines. VII. A Plane Surface is one in which, if any two points be taken, the straight line between them lies wholly in that surface. Thus the ends of an uncut cedar-pencil are plane surfaces ; but the rest of the surface of the pencil is not a plane surface, since two points may be taken in it such that the straight line joining them wUl not lie on the surface of the pencil In our introductory remarks we gave examples of a Surface, a Line, and a Point, as we know them through the evidence of the senses. 9 Book I.] DEFIMTIONS, The Surfaces, liines, and Points of (ieometry may be rej,'ar(led as mental pictures of the surfaces, lines, and points which we know from experience. It is, however, to be observed that Geometry requires us to conceive the possibility of the existence ot a Surface apart from a Solid body, of a Lino apart from a Surface. ' of a Point apart from a Line. VIII. When two straight lines meet one another, the inclina- tion of the lines to one another is called an Angle. When tim straight lines have one point common to both, they are said to f(yrm an anyle (or angles) at tlmt point. The point is called the vertex of the angle (or angles), and the lines are called the arrm of the angle (or ang)«»»\ Thus, if the lines OAy OB are terminated at the same point 0, they form an angle, which is called the angle cd 0, or the angle AOB, or the angle BOA,— the letter which marks the vertex being put between those that mark the arms. Again, if the line CO meets the line DE at a point in the line DEj so that is a point common to both lines, CO is said to make withD^ the angles COD, COE ; and th< e (as having one arm, CO, common to both) are called adjacent angles. Lastly, if the lines FG, HK cut each other in the point 0, the lines make with each other four angles FOH, HOG, GOK, KOF ; and of these GOH, F OK are called vertically opposite r.ngles^ as algo are FOff and GOI^, EUCLID'S ELEMENTS. look I. When ihne, or moi-e straight lines as OA, OB, OC, OD have a point common to all, the angle formed by one of them, OD, with OA may be regarded as being made up of the angles AOB, BOC, COD ; that is, we may speak of the angle AOD as a whole, of which the parts are the angles AOB, BOO, and COD- Hence we may regard an angle as a Magnitude, inasmuch as any angle may be regarded as being made up of parts which are themselves angles. The size of an angle depends in no way on the length of the arms by which it is bounded. We shall explain hereafter the restriction on the magnitude of angles enforced by Euclid's definition, and the important results that follow an extension of the definition. IX. When a straight line (as AB) meeting another straight line (as CD) makes the adjacent angles (ABC and ABD) equal to one another, each of the angles is called a Right Angle ; and each line is said to be a Per- pendicular to the other. C A B X. An Obtuse Angle is one which is greater than a right ansile. XI. An Acute Angle is one which is less than a right angle. XII. A Figure is that winch is enclosed by ope or mor© bpundari^s, Book I.] DKir/vn/oNS. % XI IT. A Circle is a piano figure contained by one line, whic); IS called th« Circumference, and is such, that all straight lines drawn to tho circuniferenco from a certain point (called the Centre) within the figure are equal to one another. XIV. Any straight line drawn from the centre of a circle to the circumference is called a Radius. XV. A Diameter of a circle is a straight line drawn thmugh the centre and terminated both ways by the circumference. Thus, in the diagram, is the centre of the circle ABCD, OA, OB, OC, OD are Radii of the circle, and the straight line AOD is a Diameter. Hence the radius of a circle is half the diameter. XVI. A Semicircle is the figure contained by a diametei and the part of the circumference cut of!" by the diameter. XVII. Rectilinear figures are those which are contained by straight lines. The Perimeter (or Periphery) of a rectilinear figure is the sum of its sides. XVIII. A Triangle is a plane figure contained by three straight lines. XIX. A Quadrilateral is a plane figure contained by four straight lines. XX. A Polygon is a plane figure contained by more than four straight lines. When a polygon has all its sides equal and all its angles equal it is called a regular polygon. P:ttcUD''S ELEMENTS. tfiook ' XXI. An Equilateral Triangle is one which has all its sides equal. XXII. An Isosceles Triangle is one which has two sides equal The third side is often called the 6ase of the / \ triangle. / ^ The term 6ase is applied to any one of the sides of a triangle to distinguish it from the other two, especially when they have been previously mentioned. XXIII. A Right-anoled Triangle is one in which one of the angles is a right angle. The side subtending, that is, ichich is opposite the right anale, is called the Hypotenuse. XXIV. An Obtuse-angled Triangle is /' one in which one of the angles is obtuse. ./ / It will be shewn hereafter that a triangle can have only one of its angles either equal to, or greater than, a right angle. \ XXV. An Acute-angled Triangle is one in which all the angles are acute. XXVI. Parallel Straight Lines are such as, being in the same plane, never meet when ZZZZZIZIZ continually produced in both directions. Euclid proceeds to put forward Six Postulates, or Requests, that he may be allowed to make certain assumptions on the construction of figures iinrl the properties of geometrical mag- nitudes. rUL- ook ' Book Z.] POSTULATES. Postulates Let it be granted — I. That a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. That a circle may be described from any centre at any distance from that centre. IV. That all right angles are equal to one another. Y. That two straight lines cannot enclose a space. VI. That if a straight line meet two other straight lines, 80 as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced shall at length meet upon that side, on which are the angles, which are together less than two right angles. The word rendered " Postulates " is in the original oiT^/iOTa, " requests." In the first three Postulates Euclid states the use, under certain restrictions, which he desires to make of certain in- struments for the construction of lines and circles. In Post. I. and ii. he asks for the use of the straight ruler, wherewith to draw straight lines. The restriction is, that the ruler is not supposed to be marked with divisions so as to measure lines. In Post. III. he asks for the use of a pair of compasses, wherewith to describe a circle, whose centre is at one extremity of a given line, and whose circumference passes through the other extremity of that line. The restriction is, that the compasses are not supposed to be capable of conveying distances. Post. IV. and v, refer to simple geometrical facts, which Euclid desires to take for granted. Post. VI. may, as we shall shew hereafter, be deduced from a more simple Postulate. The student must defer the consideration of this Postulate, till he has reached the 17th Proposition of Book I. Euclid next enumerates, as statements of fact, nine Axioms s EUCLms ELEMENTS. [Book L or, as he calls them, Common Notions, applicable (with the exception of the eighth) to all kinds of magnitudes, and not necessarily restricted, as are the Postulates, to geometriccU magnitudes. Axioms. I. Things which are equal to the same thing are equal to ne another. II. If equals be added to equals, the wholes are equal. III. If equals be taken from equals, the remainders are equal. IV. If equals and unequals be added together, the wholes are unequal. V. If equals be taken from unequals, or unequals from equals, the remainders are unequal. VI. Things which are double of the same thing, or of equal things, are equal to one another. VII. Things which are halves of the same thing, or of equal things, are equal to one another. VIII. Magnitudes which coincide with one another are equal to one another. IX. The whole is greater than its part. With his Common Notions Euclid takes the ground of authority, saying in effect, ** To my Postulates I request, to my Common Notions 1 olaim, your assent." Euclid develops the science of Geometry in a series of Propositions, some of which are called Theorems and the rest Problems, though Euclid himself makes no such distinction. By the name Theorem we understand a truth, capable of demonstration or proof by deduction from truths previously admitted or proved. By the name Prohlem we understand a construction, capable of being effected by the euiployment of principles of construe^ tion previously admitted or proved. A Corollary is a Theorem or Problem easily deduced from, or effected by means of, a Proposition to which it is attached. We shall divide the First Book of the Elements into three sections. The reason for this division will appear in the course of the work. 300k I] SYMBOL:^ AND ABBREVIATIONS. SYMBOLS AND ABBREVIATIONS USED IN BOOK I. •.• fw because .' therefore a is (or are) equal to L angle L triangle equilat equilateral extr. exterior intr. interior pt point rectil rectilinear © fw circle Oce circumference II parallel O parallelogram ± perpendicular reqd required rt right sq square sqq squares 8t straight It is well known that one of the chief difficulties with leumers of Euclid is to distinguish between what is assumed, or given, and what has to be proved in some of the Pro- positions. To make the distinction clearer we shall put in italics the statements of what has to be done in a Problem, and what has to be proved in a Theorem. The last line in the proof of every Proposition states, that what had to be done or proved has been done or proved. The letters q. e. f. at the end of a Problem stand for Qttod txai faciendvm. The letters q. e. d. at the end of a Theorem stand for ^itod erat demomtrandum. In the marginal references : Post, stands for Postulate. Def. Definition. Ax Axiom. I. 1 Book I. Proposition L Hyp. stands for Hypothesis, mpposition, and reftn to something jy;ntnte(l, or aBsuiiiftj to W true. fO EUCLID'S ELEMENTS. [Book r. SECTION I. On the Properties of Triangles. Proposition I. Problem. To describe an equilateral triangle on a given straigni line. Let -4 B be the given st, line. It is required to describe an equilat. A on AB With centre A and distance AB describe © BCD. With centre B and distawe BA describe © ACE. From the pt. G, in which the © s cut one another^ draw the st. lines CA, CB. Then will ABC be an equilat. A . Post. 3. Post. 3. Post. 1. For •.• ^ is the centre of © BCDj .'. AC=AB. Bet 13 And .• B is the centre of © ACE, .'. BC^AB. Del 13. Now •.• AC, BG are each= JB, .-. AC=BG. Ax. I. Thus AC, AB, BC are all equal, and an equilat. A ABC tjw b?en d??cribecl o\\ AB, Q. E. F. roo!: 1.] PkoPOsirioK II. tt PkorosiTioN II. Problem. From a given point to draw a straight line equal to a given straight lint. . -— Let A be the given pt, and BC the given st. line. It is required to draw from A a st. line equal to BC. From AtoB draw the st. line AB. Post. 1. On AB describe the equilat. A ABD. I. 1. With centre B and distance BC describe © CGH. Post. 3. Produce DB to meet the Qce CGH in G. With centre D and distance DG describe © GKL. Post. 3. Produce DA to meet the Qce GKL in L. Then will AL=BC. For •.• B is the centre of © CGEf .-. BC=BG. Def. 13. And v D is the centre of © GKL, .'. DL=DG. Def. 13. And parte of these, DA and DB, are equal. Def. 21 .'. remainder J.X= remainder BG. Ax. 3. But BC=BG ; .-. AL=BG. Ax. 1. Thus from pt. A a st. line AL has been drawn =5/^. tj. E. p. 12 EUCLID'S ELEMENTS. [Book 1 Proposition III. Problem, Frtym, the greater of two given straight lines to cut off a pa/rt equal to the less. 1.2. Ax. 1. Q. E. F. Let ABhe the greater of the two given st. lines AB, CD. It is required to cut off from AB a part = CD. From A draw the st. line AE=GD. With centre A and distance AE describe © EFH^ cutting AB in J**. ThenwiU^i'-OD. For *.• ^ is the centre of © EFH, ,\ AF=AE. Bat AE=-CD; .-. AF=GD. Thus from AB a part AF has been cut off= CD. Exercises. 1. Shew that if straight lines be drawn from A and B in the diagram of Prop. i. to the other point in which the circles intersect, another equilateral triangle will be described on AB. 2. By a construction similar to that in Prop. iii. produce the less of two given straight lines that it may be equal to the greater. 3. Draw a figure for the case in Prop, ii., in which the given point coincides with B. 4. By a similar construction to that in Prop. i. describe on a given straight line an isosceles triangle, whose equal sides shall be each equal to another given straight line. took t SookL] PR^PO<^rTTO]^ IV, ti 1.2. Proposition IV. Theorem. If two triangles have two sides of th^ one equal to two sides of the other, each to each, and have likewise the angles contained hy those sides equal to one another, they must have their third sides equal ; and the two triangles must he equal, and the other angles must he equal^ each to each, viz. those to which the equal sides are t^oaite* B C JB JP In the £.s ABC, DEF, let AB^DE, and AC^DF, and i BAC= l EDF. Then must BC^EF and A ABC = a DEF, and the other L s, to which the equal sides are opposite, must be equal, that is, I ABC-* L DEF and l AQB^ l DFE. For, if A ABG be applied to A DEF, 80 that A coincides with D, and AB falls on DE, then *.• AB=DE, .'. B will coincide with E. And *.• AB coincides with DE, and z BAC= l EDF, Hyp. .-.^0 will fall on D2^. Then •/ AC^DF, .'. C will coincide with F. And •.• B will coincide with E, and C with F, .: BC will coincide with EF ; for if not, let it fall otherwise as EOF : then the two st. lines BC, EF will enclose a space, which is impossible. Post. 5. .*. BC will coincide with and .*. is equal to EF, Ax. 8. and A ABC A DEF, and JL ABC z DEF, mil ACB z DFE. Q. B. !>• ilVcud's elements. [Book L Note 1. On the Method of Superposition. Two geometrical magnitudes are said, in accordance with Ax. VIII. to be equal, when they can be so placed that the boundaries of the one coincide with the boundaries of the other. Thus, two straight lines are equal, if they can he so placed that the points at their extremities coincide : and two angles are equal, if they can be so placed that their vertices coincide in position and their arms in direction : and two triangles are equal, if they can be so pluced that their sides coincide in direction and magnitude. In the application of the test of equality by this Method of Superposition, we assume that an angle or a triangle may be moved from one place, turned over, and put down in another place, withoat altering the relative positions of its boundaries. We also assume that if one part of a straight linb coincide with one part of another straight line, the other parts of the lines also coincide in direction ; or, that straight lines, which coincide in two points, coincide when produced. The method of Superposition enables us also to compare magnitudes of the same kind that are unequal. For example, suppose ABG and DEF to be two given angles. -B r B y Suppose the arm BC to be placed on the arm EF, and the vertex B on the vertex E. Then, if the arm BA coincide in direction with the arm ED, the angle ABO is equal to BEF. If BA fall between E;D and EF in the direction EF, ABC is less than DEF. If BA fall in the direction E<^ so that ED is between ^Q and EF, ABC is greater than DEF. Cook L] NOTE TI. IS Note 2. On the Conditions of Equality of tivo Tricmgles, A Triangle is composed of six parts, three sides and three angles. When the six parts of one triangle are equal to the six parts of another triiin^Ie, each to each, the Triangles are said to be equal in all respects. There are four cases in which Euclid proves that two tri- angles are equal in all respects ; viz., when the following parts are equal in the two triangles. 1. Two sides and the angle between them. I. 4. 2. Two angles and the side between them. I. 26. 3. The three sides of each. I. 8. 4. Two angles and the side opposite one of them. I. 26. The Propositions, in which these cases are proved, are the most important in our First Section. The first case we have proved in Prop. iv. Availing ourselves of the method of superposition, we can prove Gases 2 and 3 by a process more simple than that em- ployed by Euclid, and with the further advantage of bringing them into closer connexion with Case 1. We shall therefore give three Propositions, which we fUsignate A, B, and C, in the Place of Euclid's Props, v. vi. vri. viii. The displaced Propositions will be found on pp. 108-112. Proposition A corresponds with Euclid I. 5. B I. 26, first part I. 8. i6 EUCLID S ELEMENTS. [Book L Proposition A. Theorem. J/ two 8ide» of a triangle be eqitalj the angles opposite those sides must also he equal. Fio. 1. In the isosceles triangle ABC, let AG=AB. (Fig. 1.) Then must l ABG= i ACB Imagine the A ABC to be taken up, turned round, and set down again in a reversed position as in Fig. % and designate the angular points A\ B\ C. Then in as^BO, A'CB\ V AB=A'C, and AC=A'B\ and z BAC^ L CA'Bf, .'. L ABC= L A'CB'. I. 4. But iA'aB'=/LACB; /. z ABC= L ACB, AX. 1. Q.E.D. CoR. Hence erery equilateral triangle is also equiangular. Note. When one side of a triangle is distinguished from the other sides by being called the Base, the angular point op- posite to that side is called the Vert&j^. of the triangle. Book :.J pRorosiriox n. VI PllOrOSITION B. TlIKOIlEM. Jf two trianglen have two angles of the one equal to two angles of the other, each to each, anO the mien (idjaccnt to the equal angles in each also equal ; then must tlie triangles be equal in all respects. In A s ABC, DEF, ir« z ABC== L DEF, and ^ACB=i DFE, and BC=EF. Then must AB=DE, and AC=DF, and l BAC=^ l EDF. For if A DEF be applied to A ABC^ so that E coincides with B, and EF falls on BC ; then •/ EF=^BC, .: F will coincide with C ; and •/ 1 DEF= l ABC, .'. ED will fall on BA ; .•. D will fall on BA or BA produced. A^rain, •.• z DFE= l ACB, .'. FD will fall on CA ; .•. D will fall on CA or CA produced. .*. D must coincide with A, the only pt. common to BA and (JA. .'. DE will coincide with and .*. is equal to AB, and DF AC, and /.EDF i BAC, and (^DEF i^ABC; and .'. the triangles are equal in all respects. Q. E. D. Cor. Hence, by a process like that in Prop. A, we can prove the following theorem : If two angles of a triangle he eanal the sides which subtend ihem are also equal (^ucl. I, 6.) iS £r CUP'S ELFMEXTS. [Book 1 Proposition C. Thkorkm. If two triangles have the three sides of the one equal to the three aides of the other, each to each, the triangles must be equal in all respects. Let the three sides of the A s ABC, DEF be equal, each to each, that is, AB=DE, AC=>DF, and BC=EF. Tlien must the triangles he equal in all respects. Imajiirine the a DEF to be turned over and applied to the A ABC, in such a way that EF coincides with BC, and the vertex D fails on the side of BG opposite to the side on which A falls ; and join AD. Case I. When AD passes through BC. Then in d.ABD, V BD=BA, .-. z J?4D= z BDA, I. A. And in aACD, V GD=GA, .: l GAD= l GDA, I. A. /. sum of LsBAD, CAD=mm of i a BDA, GDA, Ax. 2. that is, z BAG^ l BDG. Hence we see, referring to the original triangles, that lBAG=- L EDF. ,'., by Prop. 4, the triangles are equal > \11 respects. [Book L Book ti I'kOPO^UTION C. ') %l to the 'le equal ml, each ed to the and the on which L, I. A. , I. A. Ax. 2. b Case II. When the line joining the vertices does not pass through BO. Then in aABD, v BD=BA, .'. l BAJ>=^ l BDA, I. A. And in aACD, '.' CD^CA, .: l QAI)^ l CDA, I. A. Hence since the whole angles BAD, BDA are equal, and parts of these CAD, CDA are equal. .-. the remainders BAC, BDC axe equal. Ax. 3. Then, as in Case I., the equality of the original triangle« may be proved. Cask III. When AC and CD are in the same struiglit line. A Then in £^ABD, v BD=BA, .*. z BAD= l BDA, I. A. that is, I IIAC= /. BDC. Then, as in Case I., the equality of the original triangles may be proved. <. E. D. - 20 EUCLID'^ ELEMENTS, Proposition IX. Problem. To bisect a given angle. £ !< a Let BAG be the given angla It is required to bisect i. BAG. In AB take any pt. D. In ^0 make AE=ADy and join DB. On DE, on the side remote from Ay describe an [iiook I 1.1. I.e. equilat.ADFE. Join AF. Then AF will bisect i BAG. For in A9AFD, AFEy •/ AD=AE, and AF is common, and FD=FE, .-. / Z)^J^= A EAF, that is, z JB^ (7 is bisected by ^i?*. .' Q. E. F. Ex. 1. Shew that we can prove this Proposition by means of Prop. IV. and Prop. A., without applying Prop. C. Ex. 2. If the equilateral triangle, employed in the construc- tion, be described with its vertex towards the given angle ; shew that there is one case in which the construction will fail, and two in which it will hold good. Note.— The line dividing an angle into two equal parts is called the BrsKCToB of the iin;jlf. Book l.]f PROPOSITION X. 21 LI. I.e. Proposition X. Problem. To bitect a ^en Jinite straight Uma, Let AB be the given st. line. It is required to bisect AB. On AB describe an equilat. A ACB. 1. 1. Bisect z ACB by the st. line CD meetin^f AB in J) ; f. 9, then AB shall be bisected in D. For in AS ACD, BCD, •.• AC=BC, and CD is common, and z ACD^ l BCD, .-. AD=BD ; L 4. .•. AB is bisected in i>. Q. E. F. Ex. 1. The straight line, drawn to bisect tht vertical angle of an isosceles triangle, also bisects the base. Ex. 2. The straight line, drawn from the vertex of an isosceles triangle to bisect the base, also bisects the vertical angle. Ex. 3. Produce a given finite straight line to a point, such that the part produced may be one-third of the line, v^hioh in piade UD of the whole and the part Droduce4r ■ ■i> 22 EUCLID'S ELEMENTS. [Book L Proposition XI. Problem. To draw a straight line at right angles to a given straight line from a given point in the same. Let AB be the given st. line, and C a given pt. in it. It is required to draw from G a st. line J. to AB. Take any pt. D in ^C, and in CB make CE=GD. On DE describe an equilat. A DFE. Join FC. FC shall be j. to AB. I. 1 For in as DCF, ECF, '.• DC=CE, and CF is common, anr' FT)=FE, .: i DCF=: A ECF ; I.e. and .-. FC is ± to AB. . Def. 9. ' Q. E. F. Cor. To draw a straight line at right angles to a given straight line JO from one extren)ity, C, take any point Din AG, produce AG to E, making CE==CD, and proceed as in the proposition. Ex. 1. Shew that in the diagram of Prop. ix. AFsmd. ED intersect each other at right angles, and that ED is bisected l.y AF. Ex. 2. If be the point in which two lines, bisecting AB «nd AG, two sides of an equilateral triangle, at right angles, ujeet ; s^ew that OA, OB, 00 are all equal. Ex. 3- Shew that Prop. xi. is a particular case of Prop, ix* [Book L ven straight ). in it. AB. D. 1.1 I.e. Def. 9. Q. E. F. to a given point D in oceed as in iFmdED ) is bisected Isecting AB •ight angles, f Prop, uc fiook i.j I^ROrO^iTiON X/J. ^i Proposition XII. Problem. To draw a straight line perpendicular to a given straight line of an unlimited length from a given jpoint without it. Let AB be the given st. line of unlimited length ; C the given pt. without it. It is rcq;uii o draw from C a st. line ± to AB. Take any pt. D on the other side of AB. With centre C and distance CD describe a © cutting Al in E and F. Bisect EF in 0, and join CE, CO, CF. I. 10 Then CO shall be ± to AB. For in as COE, COF, • •.• EO=FO, and CO is common, and CE=CF, .-. lC0E=- iCOF; I.e. .-. COis± ioAB. ' Def. 9. Q. E. p. Ex. 1. If the straight line were not of unlimited length, how might the construction fail 1 Ex. 2. If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles. Ex. 3. The lines drawn from the angular points of an equihiteral triangle to the middle points of the opposite sides ure equal. I' I it 1 V u RVCUI^S ELEMENTS. [Book t Miscellaneous Exercises on Props. I. to XII. 1. Draw a figii e for Prop. ii. for the case when the given point -4 is (o) below the line BG and t« the right of it. (/3) below the line BC and to the left of it. 2. Divide a given angle into four equal parts. 3. The angles B. G, at the base of an isosceles triangle, are bisected by the straight lines BD, GD, meeting in D ; shew that BDG is an isosceles triangle. 4. D, E, F are points taken in the sides BG, GA, JB, of an equilateral triangle, so that BD=GE^AF. Shew that the triangle DEF is equilateral. 5. In a given straight line find a point equidistant from two given points ; 1st, on the same side of it ; 2d, on opposite sides of it. 6. ABC is a triangle having the an^}eABG acute. In BA, >r BA produced, find a point D such that BD=GD. 7. The equal sides AB, AG, of an isosceles triangle ABC tre produced to points #and 0, so that AF=AG. BG and CF are joined, and H is the point of their intersection. Prove that BH=CII^ and also that the angle at A is bisected by^ff. 8. BAG, BDG are isosceles triangles, standing on oppo- site sides of the same base BG. Prove that the straight line from -4 to D bisects BG at right angles. 9. In how many directions may the line AE he drawn in Prop. III. ? • 10. The two sides of a triangle being produced, if the angles on the other side of the base be equal, shew that the triangle is isosceles. 11. ABG, ABD are two triangles on the same base AB and on the same side of it, the vertex of each triangle being outside the other. If AG=AD, shew that BG cannot =Bl>." 12. From C any point in a straight line AB, CD is drawn at right angles to AB, meeting a circle described with centre A and distance AB in D ; and from AD, AE is cut off =^0: ihew that A K8 is a right angle. [Book t he given ngle, are > ; shew , JB, of ew that Hit from opposite InBA, ;Ie ABC BG &nd Prove bisected ti oppo- ight line rawn in if the that the ise AB le being =BD. i drawn centre =AC: fiook t.] PkoPosiTIoN Xiii. 5^ Proposition XIII. Theorem. The angles which one straight line makes with another upon one side of it are either two right angles, or together equal to tivo right angles. Fig. 1. Fig. 8. JB B D Let AB make with CD upon one side of it the / s ABC, ABD. Then must these he either two rt. is, or together equal to two rt. 19 First, if I ABC^ l ABD as in Fig. 1, each of them is a rt. iC . Def . 9, Secondly, if l ABC be not= / ABD, as in Fig. 2, from J5 draw £^ ± to CD. 1.11. Then sum of z s ABC, ABD=^\\n\ oi l% BBC, EBA, ABD, and sum of z s EBC, EBD =snm of z s BBC, EBA, ABD ; .'. Slim of z s ABC, ABD=»um of z s EBC, EBD ; ■ '^' - - ■■ '-'.'■■■ Ax. 1. .'. sum of z s ABC, ABD =snm of a rt. z and a rt. z ; .'. z s ^50, ^JBD are together = two rt. z s. Q. E. D. Ex. Straight lines drawn connecting the opposite angular points of a quadrilateral figure intersect each other in <). Shew that the angles at are together equal to four right angles. Note (1.) If two angles together make up a right angle, each is called the Complement of the other. Thus, in fig. 2. z ABD is the complement of z ABE. ,il, *i two angles together make up two right anples, each IS called the Supplement of the other. Thus, in both figures, z A BD is the supplement of z ABC. ;;« ■''^ 16 RUC LID'S ELEMENTS. [Book I. PiioposiTioN XIV. Theorem. J/, ai a 2yoint in a straight line, two other straight lines, upon the opposite siiks of if., make the adjacent angles together eqnal io two right angles, these two straight lines must he in one and the same straight line. ' I. 13. Hyp. At the pt. B in the st. line AB let the st. lines BC, BD, on'opposite sides of AB, make z s ABC, ABD together = two rt. angles. Then BD must he in the same st. line udth BC. For if not, let BE be in the same st. line with BC. Then z s ABC, ABE together =two rt. / s. And I s ABC, ABD together = two rt. z s. .-. sum of z s ABC, ABE=snm of z s ABC, ABD. Take away from each of these equals the z ABC ; then z ABE= z ABD, Ax. 3. that is, the less = the greater ; which is impossible, ". BE is not in the same st. line with BC. Similarly it mny be shewn that no other line but BD is in Mie same st. line with BC. »\ BD is in the same st. line with\BO. Q. E. P. Ex. Shew the necessity of the -^ords the opposite sidest in the eniinc'.'' > i. Book I.] PROPOSITION XV. 25 Proposition XV. Theorem. // two straight lines cut one another, the vertically opposite angles must be eq^ual. Let the st. lines AB, CD cut one another in the pt. E. Then must l AEC= l BED and l AED= l BEG. For •.' AE meets CD, .'. sum of z s A EC, AED= two rt. /: a. I. 13. And •.• DE meets AB, .-. sum of z 8 BED, AED= two rt. z s ; I. 13. .'. sum of z s A EC, AED=iium of z s BED, AED ; .-. z AEC== L BED. Ax. 3. Similarly it may be shewn that z AED= z BEC. Q. K. D. Corollary I. From this it is manifest, that if two straight lines cut one another, the four angles, which they make at the point of intersection, are together equal to four right angles. Corollary II. All the angles, made by any number of straight lines meeting in one point, are together equal to four right angles. Ex. 1. Shew that the bisectors of AED and BEC are in the same straight line. Ex. 2. Prove that z AED is equal to the angle between two straight lines drawn at right angles from E to AE and EC, if both lie above CD. Ex. 3. If AB, CD bisect each other in E ; shew that the triangles ^jE^D, BEC are ec^v,\] in all respects. r, M m: flS EUCLID'S ELEMENTS. [Book L Note 3. On EiiclicPs definition of an Angle. Euclid directs us to regard an angle as the inclination of two straight lines to each other, which meet, but are not in the same straight line. Thus he does not recognise the existence of a single angle equal in magnitude to two right angles. The words printed in italics are omitted as needless, in Def. VIII., p. 3, and that definition may be extended with advantage in the following terms • — Def. Let WQE be a fixed straight line, and QP a line which revolves about the fixed point Q, and which at first coin<»*des with QK Then, when QP has reached the position represented in the diagram, we say that it has described the angle EQP. When QP has revolved so far as to coincide with QW, we say that it has described an angle equal to two right angles. Hence we may obtain an easy proof of Prop. xiii. ; for what- ever the position of PQ may be, the angles which it makes with WE are together equal to two right angles. Again, in Prop. xv. it is evident that i AED= l BECf since e«,ch has the same supplementary z A EC. We shall shew hereafter, p. 149, how this definition may be extended, so as to embrace angles greater tharf, two right g>ngle9, {Book I. Book ].] I'RorosiTio.y ,vr/. "'9 lation of •e not in ;Ie angle (lless, in ed with ^ a line at first ited in P. xqw, > right what- luakes BEC, ay be right PkOPOSITION XVI. TUKOUKM. If one side of a triangle be produced, the exterior anglt ia greater than either of the interior opposite angles. Let the side BC of a ABC be produced to D. Then must l ACD he greater than either l GAB or L ABC. Bisect ^0 in ^, and join BE. I. 10. Produce BE to F, making EF^BE, and join FG. Then in £iB BEA, FEC, :• BE=FE, and EA =EG, and l BE A = i EEC, I. 15. .-. z EGF=: L EAB. I. 4. Now lAGDia greater than z ECF ; Ax. 9. .'. z ACD is greater than z EAB, that is, z -4 CD is greater than z C4B. Similarly, if ,4 C be produced to G it may be shewn that z BGG is greater than z ^£0. and iBGG=iAGD; 1.15. .*. z -4 CD is greater than z -4£C Q. B. D. Ex. 1. From the same point there cannot be drawn mort than two equal straight lines t« meet a given straight line. Ex. 2. if, from, any point, a straight line be drawn to a given straight line making with it an acute and an obtuse angle, and if, from the same point, a perpendicular be drawn to the given line ; the perpendicul^-r wiU fall on the pide pf ihf 4 Hi 30 EUCLID'S EJJiMEMs. [Book 1 PrOPOSITIOV XVII. TlIKOUKM. Any two angles 0/ a tiiuiiyk are together leas than two righi anglei. Let ABC be any a . Then mmt any two of its la he tonether less them two rt. L s. Produce BG to D. Then /. ACD ia greater than i ABC. I. Iflt /. z s ACD, ACB are together greater than z s ABC, ACB. But JL%ACD, ACB together = two rt. i s. I. 13 .•. z a ABC, ACB are togetlier less than two rt. z s. Simihirly it may be shewn that La ABC, BA'J and alst- that I aBAC, ACB are together less than two rt. i a. Q. E. D. Note 4. On the Sixth Postulate. We learr from Prop. xvii. that if two straight lines BM and CN, which meet in -4, are met by another straight lino DE in the points 0, P, the angles MOP ana NPO are together less than two right angles. The Sixth Postulate asserts that if a line DE meeting tw. other lines n^^ CV ;m -k-s !\rOP, NTf)^ ^j^g ,^,, in'ten'or an an of fiook I.] PROPOSITIO.W xritL 3t angles on the same side of it, together less than two ri^'ht angles, £ilf and C;V sliall inuet if produced on the same side of DE on which are the angles MOt and NPO, Proposition XVIII. Theorem. If one side of a triayigle be greater than a second, the. angle opposite the first must he greater than thai opposite the second. In lABC, let side AC he greater than AB. Then must l ABC be greater than l ACB. From JO cut off ^D= J £, and join £D. Then vAB=AD, .'. iADB= lABD, - , And *.* OD, a side of A BDC, is produced to A. :. I ADB is greater than z ACB ; .'.also z ABD is greater than z ACB. Much more is z ABC greater than z ACB. 1.3. La. I. 16 Q. E. D. Ex. Shew that if two angles of a triangle be equal, the sides which subtend them are equal also (Eucl. I. 6). r 'Ul ii kUCUD'S PJJ^AfkAtTS. [Book 1. fio Proposition XIX. Theorem. If (me angle of a triangle be greater than a secondf the nide opposite the first must be greater than that opposite the second. I. A. I. 18. In i^ABC, let i ABC be greater than z ACB, Then must AG be greater than AB. For [{ AChe not greater than AB, AO must either =^^, or be less than AB. Now AC cannot =-45, for then I ABC would = z ACB, which is not the case. And AG cannot be less than AB, for then z ABC would be less than i ACB, which is not the case ; .*. ^C is greater than AB. Q. E. D. Ex. 1. In an obtuse-angled triangle, the greatest side is opposite the obtuse angle. Ex. 2. BC, the base of an isosceles triangle BAG, is pro- duced to any point D ; shew that AD is greater than AB. Ex 3. The perpendicular is the shortest straicrht line, which can be drawn from a given point to a given straieht line ; and of others, that which is nearer to the perpendicular is less than one more remote. fBook I. Book I.'. PkOPOSlTlO^ XX. 33 tcondf the nposite the 8. I. A. e. I. 18. :he case ; I. E. D. ; side is , IS prn- ?, which le ; and ■Ss than i ROPOHITION XX. TlIKOKKM. Any two »'idea of a triangle are together greater than Vie third Hdc , , - Let ^J50bea A. Then any two of its sides must be together greater than the third aide. Produce BA to D, making AD=AC, and join DC. Then v AD=AG, .'. iAGD= I ADC, that is, i BDG. I. a. Now I BCD is greater than i ACD ; .'. z BCD is also greater than i BDG ; .•. BD is greater than BG. ' I. 19. But BD^BA and AD together ; that is, BD = BA and AG together ; .'. BA and AG together are greater than BC. Similarly it may be shewn that AB and BG together are greater than AG, andBCiiudCA AB. Q. E. D, Ex. 1. Prove that any three sides of a quadrilateral figure are together greater than the fourth side. Ex. 2. .->hew that any side of a triangle is greater than the difference between the othe'* ♦wo sides. Ex. 3. Prove that the sum of the distances of any point from the angular points of a quadrilateral is greater than half the perimeter of the quadrilateral. Ex. 4. If one side of a triangle be bisected, the sum of the two other sides shidl be more than double of the line joining the vertex and the point of bisecti" n. 8. E. . •'' I 'Ml t\ 34 FMCLiD'S iiLliMENT3. [Book 1 Proposition XXI. Theorem. Jf, from the ends of the side of a trianglej there he drawn two straight lilies to a point within th$ triangle; these will he together less than the other sides of the triangle, hut mil contain a greater angle. Let ABC be a A , and from D, a pt. in the A , draw st. lines to B and 0. Then will BD, DC together he less than BA, AC, but L BDCudll be greater than l BAC. Produce BD to meet AC in K Then BA, AE are together greab«r than BE. I. 20. Add to each EG. Then BA, AC are together greater than BE, EC. Again, DE, EC are together greater than DC. I. 20. Add to each BD. Then BE, EC are together greater than BD, DC. And it has been shewn that BA, AC axe together greater than BE, EC ; .'. BA, AC are together greater than BD, DC. Next, •.• / BDC is greater than z DEC, I. i(i. and z DEC is greater than / BAC, I. IH. .; L BDC is greater than / BAC. Q. E. D. Ex. 1. Upon the base AB of a triangle ABC is described % quadrilateral figure ADEB, which is entirely within the triangle. Shew that the sides AC, CB of the triangle are together greater than the sides AD, DE, EB of the quadri- lateral. [Book t I. 20. Book I.] PROPOSTTION XXII. 35 Ex. 2. Shew that the sum of the straight lines, joinins^ the angles of a triangle with a point within the triangle, is less than the perimeter of the triangle, and greater than half the perimeter. Proposition XXII. Prohlkm, To make a triam/le, of which the sidis shall be equal to three given straight lines, any two of which are together greater than the. third. f ;{ ^i I 1. 20. I. 1(5. I. I(). Let A, B, C he the three given lines, any two of which we together g-oater than the third. It is required to make a A having its sides = A, B, C respectively. Take a st. line DE of unlimited length. j[n D£ make DF=A, FG=B, and GH==C. I. 3. With centre F and distance FD, describe ®DKL. With centre G and distance GH, describe ®HKL. Join FK and GK. Then aKFG has its sides =A, B, (7 respectively. For FK=FD ; Def. 13 .-. FK=A; and GK = GII; Def. 13. .-. GK = C; and FG = B; .'. a l^KFG has been described as reqd. q. e. f. Ex. Draw an isosceles triangle having each of the equ^l f|d^9 double of the bas^, '1*1 m 36 EUCLID'S FT FMENTS. [Book I. PiioposiTTON XXIII. Problem. At a given point in a given straight line, to make an angle equal to a given angle. Let A be the given pt., BC the given line, DBF the given z . It is reqd. to make at pt. A an angle = z DBF. In ED, EF take any pts. D. F ; and join DF. In AB, produced if necessary, make AG=DE. In AC, produced if necessary, ninke AH=EF. In HC, produced if necessary, make HK=FD. With centre A, iind distance AG, describe ® OLM. With centre if, and distance HK, describe ® LKM. Join ^L and HL. Then •.• LA = A G, .: LA =DE ; Ax. 1 and •.• HL=HK, .-. HL=FD. Ax. 1. Then in lb LAH, DEF, '.• LA=DE, and AH=EF, and HL=FD ; .-. z LAH= L DEF. I. c. ff fl.n angle LAH has been made at pt. A as was reqd. 9' »f ?• Book tj PROPOSITION P>. 37 Note. — We here give the proof of v, theorem, necessary to the proof of Prop. XXIV. and applicable to several proposi- tions in Book III. r\ Proposition D. Theorem. Every straight line, drawn from the vertex of a triangle to the hose, is less than the greater of the tivo sides, or than either, if they be equal. In the LABCi let the side AC ho not less than AB. Take any pt. D in BC, and join AD. Then must AD he less than AG. For •.* AC is not less than AB ; .-. ^ ^BD is not less than i AGD. I. a. and 18. But I ADC is greater than z ABD ; I. 16. .•. z ADC is greater than i A CD ; .'.AC la greater than AD. I. 19. Q. B. p. 38 k UCUP S EL EMkNTS. [Book t. Proposition XXIV. Theorem. If two triangles have two sides of the one equal to tivo aides of the other, each to each, hut the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other ; the base of that ivhich has the greater angle must be greater than the base of the other. In the A8 ABC, DBF, let AB=DE And AC =DF, and let z BAC be greater than z EDF. Then must BC be greater than EF. Of the two sides BE, DF let BE be not greater than DF* At pt. D in St. line ED make z EDG= z BAC, I. 23, and make DG=AC or DF, and join EG, GF. Then'.' AB='DE, and AC=DG, and z BAC= z EDG, ,'. BC=EG, I. 4. Again, •.' DG=DF, :. lDFG^iDGF', La. .'. z EFG is greater than z DGF ; much more then z EFG is greater than z EGF ; .'. EG is greater than EF. I. 10. Bnt EG^BG; .: BC is greater than EF. Q. E. D. •This line was added by Simson to obviate a defect in Euclid's proof. Without tliis condition, three distinct cases irinst be discussed. With the condition, we can pi'ove that F must lie below EO. For since DF is not less than DE, and DO is drawn equal to DF, DG is not less than DE. Hence by Prop. D, any line drawn from D to' meet EO is less than DG, and therefore DF, being equal to DO, must extend beyond EG. For another method of proving the Proposition, see p. 113. jBOOK lj PROPOSITION XXV, S 39 »■'■.! 1 1. A. I. in. Proposition XXV. Theorem. Ij two triangles have two sides of the one equal to two sides of the other, each to each, hut the base of the one greater than the base of the other ; the angle also, contained by the sides of that which has the greater base, must he greater than the anglo contained by the sides equal to them of the other. In the LsABC,DEF, let AB=DE and AC=DF, and let BG be greater than EF. Then must l BAC he greater than l EDF. For z BAG IB greater than, equal to, or less than z EDF. Now I BAG csmnot^ l EDF, for then, by i. 4, BG woyx\d= EF ; which is not the case. And z BAG cannot be less than z EDF, for then, by i. 24, BG would be less than EF ; which is not the case ; .*. z BAG must be greater than z EDF. Q. E. D. Note. — In Prop. xxvi. Euclid includes two cases, in which two triangles are equal in all respects ; viz., when the following parts are equal in the two triangles : 1. Two angles and the side between them. 2. Two angles and the side opposite one of them. Of these we have already proved the first case, in Prop, b, 80 that ^ have only the second case left, to form the subject of Prop. XXVI., which we shall prove by the method of superposition. For Euclid's proof of Prop, xxvi , see pj. 114-1 16. r on the same side of the line, or male the interior angles upon the same side together equal to two right angles ; the two straight lines are parallel to one another. I. Hyp. I. 15. II. Let the st. line EF, falling on st. lines AB, CD, make I. z J5<'6rjB = corresponding z (?/fi), or II. z 8 BGH, GHD together=two rt. z s. Then, in either case, AB must be \\ to CD. V I EGB m given = I GHD, and z EGB is known to be= z AGH, .: lAGH= lGHD', and these are alternate z s ; .-. AB is II to CD. V I 8 BGH, GHD together = two rt. z s, and z s BGH, AGH together = two rt. z s, /. L% BGH, AGH together = z 8 BGH, GHD together ; /. z AGH=- L GHD } .-. AB is II to CD. L 27. 1.27. Hyp. I. 13 Q. B. D. Hyp. I. 15. 1.27. Hyp. I. 13. her; Book I.] i\OTE v. OX THE SIXTH rOSTCl.A'li:. 47 Note 5. 0/t the- Sixth Postulate. In the place of Euclid's Sixth Postulate ninny modern Dvriters on Geometry [>ropo.se, us more evident to the aenses, the following Postulate : — " Two straight lines which cut one anuthrr cannot both he paralfel to the same straight line." If this be assumed, we can prove Post. 6, as a Theorem, (Ims : Let the line EF falling on the lines AB, CD make the z s hGH, OIID together less than two rt. i a. Then uuist AB^ VD meet when produced towards B^ D. For if not, suppose AB and CD to be parallel. Then •.• z s AGH, JBG/f togcther=two rt. z s, I. 13. and I s GHD, BGH are together less ,than two rt. l s, .'. L AG^ is greater than z GHD. Make z MGH= z GHD, and produce MG to N. Then *,; the alternate z s MGH, GHD are equal, .-. Mi\ is II to CD. I. 27. Thus two lines MN, li which cut one another are both parallel to ( 'D, which is impossible. .'. AB and CD are i.jt parallel. It is also clear that they meet towaids B, D, because 67' lies between GN and HD, Q. K. D HI ■ u n; i < ''I J 'if ' ■ h )» .11 ■ti 48 EUCLID'S ELEMENTS. [Book lo f» Proposition XXIX. Theorem. // a straight line fall upon two piO'Talld straight lines, it makes the two interior angles upon the same side together equal to two right angles, and also the alternate angles equal to one another, and aXao the exterior angle equal to the interior and opposite upon the same side. ^ ^^' Let the st. line EF fall on the parallel st. lines AB, CD. Then mv^t I. z s BGH, GHD together = two rt. z s. II. z ^G«if= alternate z GHD. III. z iJG'J5= corresponding z GHD. I. z s BGH, GHD cannot be together less than two rt. z s, for then AB and CD would meet if produced towards B and D, Post. 6. which cannot be, for they are parallel. Nor can z s BGH, GHD be together greater than two rt. z s, for then z s AGH, GHC would be together less than two rt. z 8, I. 13. and AB, CD would meet if produced towards A and C Post. 6 which cannot be, for they are parallel, .-. z s BGH, GHD together = two rt. ^ s. :. '.' L s BGH, GHD together = two rt. z s, and z s BGH, AGH together = two rt. z s, I. 13. .-. z 8 BGH, AGH together= z s BGH, GHD together, and .-. z AGH^ z GHD. UI. V lAGS^^lGHD, -^d z AGH= L EGB, .-. z EGB= L GHD, Ax. 1.15. Ax. 1 Q. E. D. Sook L] PROPOSITION XXIX. 49 £XERCISE& 1. If through a point, equidistant from two parallel straight lines, two straight lines be drawn cutting the parallel straight lines ; they will intercept equal portions of the parallel lines. 2. If a straight line be drawn, bisecting one of the angles of a triangle, to meet the opposite side ; the straight lines drawn from the point of section, parallel to the other sides and terminated by those sides, will be equal. 3. If any straigh^ine joining two parallel straight lines be bisected, any other straight line, drawn through the point of bisection to meet the two lines, will be bisected in that point. Note. One Theorem (A) is said to be the converse of another Theorem (B), when the hypothesis in (A) is the concluadon in ^3), and the conclusion in (A) is the hypothesis in (B). For example, the Theorem I. a. may be stated thus : Hypothem. If two sides of a triangle be equaL Conclusion, The angles opposite those sides must also be equal The converse of this is the Theorem I. b. Cor. : Hypothesis. If two angles of a triangle be equal Conclusion. The sides opposite those angles must also be equal. The following are other instances ; Postulate VI. is the converse of I. 17, I. 29 is the converge of I. 27 and 28. SI. i! & s. «^ 50 EUCLID'S ELEMENTS. [Book L Proposition XXX. Theorem. Straight lines which are parallel to the earns straight Iw" are par aUd to (me another. Let the st. lines AB, CD be each || to EF. Then must AB he \\ to CD. Draw the st. line Crif, cutting AB, CD, EF in the pts. Then •.• QH cuts the || lines AB, EF, .-. /. ^OP=alternate l PQF. And •.• GH cuts the || lines CD, EF, .: extr. z OPD= intr. z PQF , .'. L AOP= L OPD ; and these are alternate angles ; .-. AB is II to CD. 1.29. 1.29. 1.27. Q. E. D. The following Theorems are important. They admit of easy proof, and are therefore left as Exercises for the student. 1. If two straight lines be parallel to two other straight lines, each to each, the first pair make the same angles with one another as the second. 2. If two straight lines be perpendicular to two other straight lines, each to each, the first pair make the same angles with one another as the second, dook t] PROPOSITTON XXX f. St Proposition XXXI. Problem. , To draw a sbaiyht line through a given point parallel */0 a given straight line. E J' m n ^ ~u Let A be the given pt. and BC the given st. line. It is required to draw through A a st. line \\ to BC. In BC take any pt. D, and join AD. Make i DAE= i ADC. I. 23. Produce EA to F. Then EF shall be ;| to BC. For •.• AD, meeting EF and BC, makes the alternat«» angles equal, that is, z EAD= l ADC, .'. EF ia\\ to BC. 1.2' .*. a St. line has been drawn through ^ |l to BC^ Q. E. F. Ex. 1. From a given point draw a straight line, to make '^ an angle with a given straight line that siuiU be equal to a given angle. , Ex. 2. Through a given point A draw a straight line ABC, meeting two parallel straight lines in B and C, so that BC may be equal to a given straight line. % Q7 -h .'-^ H-njjCUL'r 7"/" ■' ■vX— ' /•'u-eo-i r\ I' '?l- i./t c 5* EUCLID'S ELEMENTS. tBook L , -"'" Proposition XXXII. Theorem. If a side of any triangle be produced, the exterior angle is equal to the tiro interior and opposite angles, and the three interior angles of every triangle are together equal to two right angles. 1.31. 1.29. I. 29. ^ Let ABQ be a A , and let one of its sides, 50, be pro- duced to D. Then will l^- L z ACD= L s ABC, BAC together. II. z s ABC, BAC, ACB together =two rt is. From C draw CE \\ to AB. Then I. •.• BD meets the lis EC, AB, .'. extr. z ECD={r\ir. lABC. And •.• -40 meets the ||s EC, AB, .-. I -40^= alternate ^ BAC. .-. L s ECD, ACE together= z s ABC, BAC together ; .'. iACD=/.sABC,BAC together. And II. •.' IS ABC, BAC together= z ACD, to each of these equals add i ACB ; then z s ABC, BAC, ACB together = is ACD, ACB together, .-. z s ABC, BAC, ACB together = two rt. z s. I. 13. Q. E. D. Ex. 1. In an acute-angled triangle, any two angles are greater than the third. Ex. 2. The straight line, which bisects the external vertical angle of an isosceles triangle is parallel to the base. th( Book I.] PROPOSITION XXXII. S3 Ex. 3 If the side BO of the triangle ABC be produced to D, and AE be drawn bisecting the angle BAG and meeting BQ in B ; shew that the angles ABD, ACD are together double of the angle A ED. Ex. 4. If the straight lines bisecting the angles at the base of an isosceles triangle be produced to meet ; shew that they will contain an angle equal to an exterior angle at the base of the triangle. Ex. 5. If the straight line bisecting the external angle of a triangle be parallel to the base ; prove that the triangle is isosceles. The following Corollaries to Prop. 32 were first given in Simson's Edition of Euclid. Cor. 1. The sum of the interior angles of any rectilinear figure together with four right angles is equal to twice as many right angles as the figure lias sides. U til ■i'l' m m Let ABODE be any rectilinear figure. Take any pt. F within the figure, and from F draw the St. lines FA, FB, FO, FD, FE to the angular pts. of the figure Then there are formed as many z s as the figure has sides. The three z s in each of these As together = two rt. z s. .'.all the zs in these as together = twice as many right z s as there are A s, that is, twice as many rii'ht z s as the figure has sides. ' Now angles of all the As= z s at A, B, 0, L, E and l s SitF, that is, = z s of the figure and ^a at F, and .'. =s= z 8 of the figure and four rt. z s. I. 15. Cor. 2 .*. z s of the figure and four rt. z s = twice as many rt ^ § as the figure has sicjes, 54 EUCLID'S ELEMENTS. [Book I. Cor. 2. The, exterior angles of any convex rectilinear figure, made by producing each of its sides in succession, are together equal to four right angles. Eveiy interior angle, as ABC, iind its adjacent exterior angle, as ABD, together are = two rt. z s. .'. all the intr. z s together with all the extr. z s = twice as many rt. z s as the figure has sides. But all the intr. / s together with four rt. z 8 ■= twice as many rt. z s as the figure has sides. .', all the intr. i s together with all the extr. 4. a =all the intr. z s together with four rt. z s. .*. all the extr. z s = four rt. z s. Note. The latter of these corollaries refers only to convey figures, that is, figures in which every interior angle is less than two right angles. When a figure contains an angle greater P 81 i than two right angles, as the angle marked by the dotted line in the diagram, this is called a reflex angle. See p. 149. Ex. 1. The exterior angles of a quadrilateral made by pro- ducing the sides successively are together equal *o the interior {i*»gle8. ; fiook I.] rko POSITION XXX II I. • » 13 Ex. 2. Prove that the interior angles of a hexagon are equal to eight right angles. Ex. 3. Shew that the angle of an equiangular pentagon is ^ !'f a right angle. Ex. 4. How many sides has the rectilinear figure, the sum of whose interior angles is double that of its exterior angles ? Ex. 6. How many sides has an equiangular polygon, four of whose angles are together equal to seven right angles ? Proposition XXXIII. Theorem. The, straight lines which join the extremities of tiro equal and parallel straight lines, towards the same parts, are also them- selves equal and parallel. Let the equal and || st. lines AB, CD be joined towards tb* sjime parts by the St. lines J^ 6', £Z>. Then must AG and BD he equal and ||. Join BG. Then .• AB i»\\ to CD. .-. z JiJ(7= alternate l DCB. 1.29. Then in As ABC, BCD, '> ••• AB=GD, and BC is common, and z ABC= i DCB, ■ .'.AG= BD, and i ACB= i J^BC. I. 4. Tb«u ••• BC, meeting AC and BD, makes the alternate z s ACB, DI>^ equal, .'.AGisWtoBD. (i. SI. D. "^i \\ m M 5« ^VCtWS ELEMENTS. [Book L 1. Miscellaneous Exercises on Sections 1. and II. If two exterior angles of a triangle be bisected by t V straight lines which meet in ; prove that the perpend iculara from on the sides, or the sides produced, of the triangle are equal. 2. Trisect a right angle. 3. The bisectors of the three angles of a triangle meet in one point. 4. The perpendiculars to the three sides of a triangle drawn from the middle points of the sides meet in one point. 6. The angle between the bisector of the angle BAC of the triangle ABC and the perpendicular from A on BC, is equal to half the difference between the angles at B and C. 6. If the straight line AD bisect the angle at A of the triangle ABC, and BDE be drawn perpendicular to AD, and meeting ^C, ot AC produced, in E; shew that BD ia equal toD^. 7. Divide a right-angled triangle into two isosceles tri- angles. 8. AB, CD are two given straight lines. Through a point E between them draw a straight line GEH, such that the in- tercepted portion GH sliull be bisected in E. 9. The vertical angle of a triangle OPQ is a right, acute, or obtuse angle, according as OR, the line bisecting PQ, is equal to, greater or less than the half of PQ. 10. Shew by means of Ex. 9 how to draw a perpen- dicular to a given strf«ight line from its extremity without pro- ducing it. 1 Y Book I.] MVcUlys ELEMENn, S7 SECTION III. On the Equality of Rectilinear Figures in respect of Area. m 1 1 The amount of space enclosed by a Figure is called the Area of that fifrure. Euclid calls two figures tqml when they enclose the same amount of space. They may he dissimilar in shape, but if the areas contained within the boundaries of the figures be the same, then he calls the figures t^uxd. He regards a triangle, for example, iis a figure having sides and angles and area, and he proves in this section that two triangles may have equality of area, though the sides and angles of each may be unequal. Coincidence of their boundaries is a test of the equality of all geometrical magnitudes, as we explained in Note 1, page 14. In the case of lines and angles it is the only test : in the case of figures it b a test^ hut not the only test ; as we shall shew in this Section. The sign —, standing between the symbols denoting two f^v/reSf must be read is equal in area to. Before we proceed to prove the Propositions included in this Section, we must complete the list of Definitions required in Book I., continuing the numbers prefixed to the definitions in page 6. 55 KVCLlD'S ELEMENTS. [Book 1. Definitions. XXVII. A Parallelogram ia a four-sided figure whose opposite sides are parallel. For brevity we often designate a parallelogram by two letters only, which mark opposite angles. Thus we call the figure in the margin the parallelogram AQ. XXVIII. A Rectangle is a par- allelogram, having one of its angles a right angle. Hence by I. 29, all the angles of a rectangle are right angles. XXIX. A Rhombus is a par- allelogram, having its sides equal. XXX. A Square is a paral- lelogram, having its sides t'qiial and one of its angles a right angle. Hence, by I. 29, all the angles of a square are right angles. XXXI. A Trapezium is a four-sided figure of which two sides only are parallel. XXXII. A Diagonal of a four-sided figure is the straight line joining two of the oppisiie nnfru'nr }> infs, >ok 1. Book I.] EXERCISES ON DEFINITIONS 27-33. 59 7 two tlie Ight Tht ht XXXIII. The Altitude of a Parallelogram is the perpen- dicular di»tance of one of its sides from the side opposite, regarded as the Base. The altitude of a triangle is the perpendicular distance of one of its angular points from the side opposite, regarded as the base. Thus if ABCD be a parallelogram, and AE a perpendicular let fall from A to CD, AE is the altitude of the parallelogram, and also of the triangle -4 CD. If a perpendicular be let fall from B to DC produced, meet- ing DC* in F, BF is the altitude of the parallelogram. Exercises. Prove the following theorems : 1. The diagonals of a square make with each of the sides •**) an^rle equal to half a right angle. 2. If two straigiit lines bisect each other, the lines joining theii; extremities will form a parallelogram. 3. Straight lines bisecting two adjacent angles of a paral- lelogram intersect at right anoles. 4. If the straight lines joining two opposite angular points of a parallelogram bisect the angles, the parallelogram has all its sides equal. 5. If the opposite angles of a quadrilateral be equal, the (jiuadrilateral is a parallelogram. 6. If two opposite sides of a quadrilateral figure be equal to one another, and the two remaining sides be also equal to one another, the figure is a parallelogram. 7. If one angle of a rhombus be equal to two-thirds of two right angles, the diagonal drawn from that angular point divides the rhombus into two equilateral triangles, f'-i r I 'CI. m'S EI. EM EM'S, [Book L Proposition XXXIV. Tiikoiiem. " " The opponite sides ami angles of a parallelogram are equal to nyie another, and the diagoiial bisects it. Let ABDC be a O, and BC a diagonal of the O. Then must AB=:DC &nd AC^DB, and L BAG-^ l CDB, and i ABD^ l ACD ' and lABC= lDCB. For vABi^W to CD, and BC meets them, .*. z 2lB0= alternate i DCB , i 2& and •.• ^C is II to BD, and BC meets them, .-. z ^0£=alternate z D^O. ^ 1.29. Then in l^ ABC, DCB, '.' z ABC=- L DCB, and l ACB^ l JfVC, and BC is common, a side adjacent to the equal z s in each ; .-. AB^DC, and AC= DB, and z £J (7= z CDB, and A ABC= A DOB. Lb. Also ••• z ^2JC= z DCB, and z DBC= z ^ (7B, .-. z s ABC, DBC together = z s DCB, ACB together, that is, iABD=^iACD. Q. E. D. Ex. 1. Shew that the diagonals of a parallelogram bisect each other. !^x, % Shew th^it the diagonals of a rectangle are ec^^ual. ' ok L bOOK I.] PROPOSITION' XXXV. 61 il to id Proposition XXXV. Theorem. Parallelogra/ms on the same hose and between the same pwalleh are equal. Let the Os ABCB, EBCF be on the same base BC and between the same ||s AF^ BC. Then must O ABCD-CJ EBCF. Case I. If ADy EF have no point common to both) Then in the AS FDC, i;^B, •••extr. LFDC^iniv. i EAB, and intT. I DFC=extT. /.AEB, mdDC=AB, /. lFI)C= lEAB. Now O ABCD with A i^DC=figure ABCF ; and O EBCF with A ^.4B= figure ABCF\ .'. O ABCD with A FDC=CJ EBCF with A EAE-, .'. O ABCD^CJ EBCF. 1.29. I. 29. 1.34. St Case IL If the sides AD, EF overlap one another the same method of proof applies* EUCLID'S ELEMENTS. [Book ), Case III. If the sides opposite to BC be terminated in the same point />, 42- 1.34. 3 V the same method of proof is applicable, but it is easier to reason thus : Each of the Os is double of A BDV. ; .-. O ABCD=CJ DBCF. Q. B. D. Proposition XXXVI. Theorem. Parallelograms on equal hasesy and between (he same paralkUf are equal to one another. A. .o !■: n S T,et the Os ABGD, EFGR be on equal bases BC, FGy and between the same \\s AH, BG. Then must O ABCD=EJ EFGH. u uin BE, CH. Then v BC=FG, and EII=FG; .\BC=EH', mdBCis\\toEH. /. EB is II to CH ; .*. EBCH is a parallelogram. Now O EBCH^ EJABCD, V they are on the same base BC and between the z'-vuie ||s ; and O EBCH = C7EFGH, I . .io. V they are on the same base EH and between the same j|8 , .\CJAr.rr)=ULF(UL Q. m, D Hyp. 1.34. Hyp. I. 33. 1.35. :ook I.] PROPOSITIOmV XXXVlt. ^ Proposition XXXVII. Theorem. 'J'lKinyles Kjwn the same base, and between the sama parallels, are equal to one another. Let A s ABC, DBC be on the same base BC and between the same Ijs AD, BC. Then must A ABC= a DBC. Frdu B draw BE H to CA to meet DA produced in E. From C draw CF || lo BD to meet AD produced in F. Then EBCA and FCBD are parallelograms, und nJEBGA=CJFCBD, 1.35. '.' they are on the same base and between the same ||s. Now A A BC is half of CJ EBCA, I. 34- and A DBC is half of O FCBD ; I. 34. .-. aABC=aDBC. Ax. 7. Q. E. D. 1*1 X. 1. If P be a point in a side AB of a paralleloorram ^ A nCD, and PC, PD be joined, the triangles PAD, PBC are together equal to the triangle PDC. „^ Ex. 2. If A, B be points in one, and C, D points In / another of two pnnillel straight lines, and the lines AD, BC inte.sei't in I], then ilw triangles AE(\ BED nre rqiuil. ml I MVCLWS kLEMBNTS, [Bo*«m Proposition XXXVIII. Theorem. Triangles upon equal haseSf and between the same paralleh, are equal to one another. E B jr j{ Let AS ABCj DEF be on equal bases, J30, EF^ and between the same ||s BF^ AD. Then must A ABC^ lBEF. From B draw BO || to CA to meet DA produced in O. From F draw FH U to ED to meet AD produced in R. Then CO and EH are parallelograms, and they are equal, *.* they are on equal bases £(7, EF, and between the same l^BFyOH. 1.36 Now A -4BC7 is half of O C(?, Mid A DJ^J* is half of O -fifl^ ; .'. lABC^lDEF. Ax. 7. Q. E. D. Ex. 1. Shew ^hat a straight line, drawn from the vertex of a triangle to bisect the base, divides the triangle into two equal parts. Ex.2. In the equal sides ABf AG oi an isosceles triangle ABC points D, E are taken such that BD^AE. Shew that the triangles CBD, ABE are equal. i; Book L] PROPOSITION XXXIX. K tu Proposition XXXIX. Theorem. Eqml triangles upon the same base, and upon the sa/me nidn. of it, a/re between the same parallels. 'I I li; •! ■; u Let the equal a b ABC, DBC be on the same base BGj and on the same side of it. Join AD. Then must AD he || to BG. For if not, through A draw A \\ to BC, so as to meet BD, 01 BD produced, in 0, and join OC. Then *.* as ABC, OBC axe on the same base and between the same ||s, « .'. i^ABC= c^OBC. 1.37. But lABG=lDBC', Hyp. .-. lOBG^ lDBG, the IeBSaralleloijram, is double of the triangle. m ! f u 1^ Let the O ABCD and the a EBC be on the same base BG and between tlie satne ||s AE, BV. Then must O ABCD be double of A EBC, Join AG. Then lABG= a EBC, v they are on the same base and between the same II 8 ; L 37. and O ABCD is double of A ABO, v AGis a diagonal of ABCD ; L 34. .-. O ABCD is double oi A EBC. Q. E. D. Ex. 1. If from a point, without a parallelogram, there be drawn two straight lines to the extremities of the two opposite sides, between which, when produced, the point does not lie, •-T" the difference of the triangles thus formed is equal to half the parallelogram. Ex. 2. The two triangles, formed by drawing straight lines . from ai.y point within a parallelogram to the extremities of its opposite sides, are together half of the parallelogramf V- 68 EUCLID'S ELEMEXTS, [Book X. Book < Proposition XLII. Problem. To describe a paralliloiiraiii that shall be equal to a given triangle, and have one of its amjle-i equal tu a given angle. / Let A BC be the given A , and D the given z . It is required to describe a CO equal to A ABC, having one of its Ls= L D. Bisect BC in E and join AE. I. 10. At ^ make i CEF= l JD. I. 23. Draw AFG\\ to BC, and from C draw CG\\ to EF. Then FECG is a parallelogram. Yiovf lAEB=lAEC, '.' they are on equal bases and between the same lis. I. 38. .-. A ^jBO is double of A ^^0, But O FECG is double of a AEC, '.' they are on same base and between same ||s. I. 41. .-. O FECG^ a ABC ', Ax. 6. and nj FECG has one of its z s, CEF= z D. .*. O FECG has been described as was reqd. Q. E. F. Ex. 1. Describe a triangle, which shall be equal to a given parallelogram, and have one o^ its angles equal to a given rectilineal angle. Ex. 2. Construct a parallelogi-am, equal to a given triangle, and such that the sum of its sides shall be equal to the sum of the sides of the triangle. Ex. 3. The perimeter of an isosceles triangle is greater than the perimeter of a rectangle, which is of the same altitude Ijrithf pud ec[ual to, the given triangle, Jabci ^ and t ^X in the Book I.] pkOPosiTioN xun. (-^J PROPOSTTTON XLIII. THEOREM. The complements of the iiaraUehvjrmm^ which ore about the diameter of any 'parallelogram, are equal to one another. Let ABCD be a O, of which BD is a diagonal, and EG, HK the CJh about BD, that is, through which BD passes, and AF, FC the other Os, which make up the whole figure ABCp, and which are .'. called the Complements. Theyi must comjilement AF=comiilement FG. ' For •.• BD is a diagonal of CJ AG, .-. A ABD= A GDB ; I. 34. and •.' BF is a diagonal of ZZZ HK, .: A HBF=aKFB', 1.34. and •.' FD is a diagonal of O EG, .'. A EFD= A GDF I. 34. Hence sum of as HBF, EFD ==snm of as KFB, GDF. Take these equals from as ABD, GDB respectively, then remaining O ^i''= remaining O FC. Ax. 3. Q. E. D. ' y Ex. 1. if through a point Q, . within a parallelogram \J A BCD, two straight lines are drawn parallel to the sides, ^/ and the parallelograms OB, OD are equal j the point is in the diagonal AG. Ex. 2, ABCD is a parallelogram, AMN a straight line meeting the sides BG, CD (one of them being produced) iti M, N. Shew that the triangle MBN is equal to the trian-'o MDO, ■■o V 4r I' 1 I 76 i:VCLIiys ELEMENTS. [Book L Proposition XLIV. Problem. To a given straight line to apphf a parallelogram, which shall he equal to a given triangle, and have one of its angles equal to a given angle. Let AB be the given st. line, C the given A, D the iriven z . It is required to apply to AB a HJ — A C and having eve of its I s= L D. Make a ZI7= A C, ai|d having one of its angles = i D, I. 42. iiiid suppose it to be removed to such a position that one of tlie sides containing this angle is in tlie sjinie st. line with AB, and let the £7 be depleted by BEFG. Produce FG to H, draw All \\ to BQ or EF, and join BIT. Then •/ FH meets the [js AH, EF, .'. sum of z s J TTF, ITFE=t\vo rt. z s ; I. 20. .'. sum of z s BHG, HFE is less than two rt. z s ; .•. HB, FE will meet if produced towards B, E. Post. 6- Let them meet in K. Through K draw KL || to EA or FH, J~ "i V and produce HA, GB to meet KJj in the pts. L, M. Then HFKL is a O, and HK is its diagonal ; and AG, ME are Os about HK, .*. complement 5Z(= complement JBi^, 1,42 .: C7BL= aC. Also the CJ BL has one of its z s, ABM^ L EBb,', and .*. equal to z D. !•:. V [Book L Book I.] PROPOSITION XL P: n I, which 8 angles D the ling eve. , I. 42. t one of ith AB, bin BTL 1.20. zs; Post. 6 L M 36r', and Proposition XLV. Problem. To describe a parallelogram, which shall he equal to a given rectUinear Ji'/xre., and have one of its angles equal to a given angle. Let ABCD be the triven rectil. fij^ure, and E the given Z . It is required to describe a CJ = to ABCD, having one of its 18= I E. Join AO. Describe a O FGHK= A ABG, having z FKH= /. E. I. 42. To C?ff apply a O GHML= a 01)^, bavin^r ^ r^//M= z £?. I. 44. Then JP'jrML is the O req.l. For •.• z GHMmd z FKH Are each= z ii' ; • .-. AGHM=iFKH, .-. sum of z s OTiM, GHK=min of z s i^'ilTF, GHK = two rt. z s ; I. 29. .*. KHJ\£ is a st. line. I. 14. Again, '.• HG meets the ||s FG, KM, lFGH^lGHM, .'. sum of z.s FGH, LGH =aiim of z s GHM, LGH =two rt. z 8 ; I. 29. .*. FGL is a st. line. Then •.' KF is || to HG, and H(? is I| to LM .-. iiCiJ' is II to LM ; and KM has been shewn to be || to FL, .'. FKML is a parallelogram, and •.• Fn= a ABC, and GM= a Oi)^, .-. £7 i^ilf=- whole rectil. fig. ABCD, and O i^iVf has one of its z s, FKM= z ^. In the same way a CJ may be constructed equal to a given rectil. fig. of any number of sides, and having one of its angles equal to a given angle. Q i" v. I. 14. ISO. 11 .!i :'| J ya EUCLID'S ELEMENTS. [Book L Miscellaneous Exercisett Bo( 1. Ir one diagonal of a quadrilateral bisect the other, it divides the quadrilateral into two oqual triangles. 2. If from any point in the dingonal, or the dingonal pro- duced, of a parallelogram, straight lines be drawn to the opposite angles, they will cut o£f equal triangles. 3. In a trapezium the straight line, joinirig the middle points of the parallel sides, bisects the trapezium. 4. The diagonals AG, BD i»f a parallelogram intersect in 0, and P is a point within the triangla.-405 ; prove that the difference of the triangles CPD, APD' ia equal to the sum of the triangles APC, BFD. 5. If either diagonal of a parallelogram be equal to a side of the figure, the other diagonal shall be greater than any side of the figure. 6. If through the angles of a parallelogram four straight lines be drawn parallel to its diagonals, another parallelogram will be formed, the area of which will be double that of the original parallelogram. 7. If two triangles have two sides respectively equ.il and the included angles supplemental, the triangles are equal. 8. Bisect a given triangle by a straight line drawn from ii given point in one of the sides. 9. The base AB of a triangle ABC is produced to a point D such that BD is equal to AB, aua straight lines are drawn from A and T) to E, the middle point of BC ; prove that the triangle ADE is equal to the triangle ABC. 1(». Prove that a pair of tlie di iv" ils ol ihe parallelograms, w hich fire about the diameter of any para'lelogram, are paralW to eadi oil ' ", eq eq in Book I.] pRuroiirnoN xwl 73 ^ i Proposition XLVI. Problem. To deserihe a square vpon a given atra/ight Um. Let AB be the given st. line. Jt i» required to describe a square on AJB» From A draw ^C ± to AB. In AC make AD=AB. Through D draw DE II to AB. Through B draw BE || to AD. Then AE is a parallelogram, and . •. AB^ED, and AD=BE. Bui A^^ AD; .'. AB, BE, ED, DA are all equal ; .'. AE is equilateral. And z BAD is a right angle. .*. AE is a square, and it is described on AB. 1.34. Def. XXX. Q. B. p. Ex. 1. Shew how to construct a rectangle whose sides are equal to two given straight lines. Ex. 2. Shew that the squares on equal straight lines are equal. Ex. 3. Shew that equal squares must be on equal straight lines. Note. The theorems in Ex. 9 and 3 are mwumed by Eudid in the proof ol Prop. XLVin. m L 11. Cor. I. 31. 1.31. 1 i 74 EUCLID'S ELEMEXTS. Book L /^ Proposition XLVII. Theorem. In any rightanfih'.d frlniHift'. tin: square tvhich is described on the side siibtendiug the rii/ht (tiujh is equal to the squares described on t\c sides which coutaiit the right angle. Let ABC be a rifrht-analed A , having the rt. z BA 0. Then vnnd xq. on HC=^ sum of sqq. on BA, AO. On BC, CA, A ti .Usci. the sqq. BDE(\ CKHA, AOFB. Througli A draw AL \\ to BU or CE, and join ADy FC. Then •.' z BAG and z BAG are both rt. z s, .*. CAG in a St. line ; I. 14 and •.• z BAG and z CAH are both rt. Z a ; .'. HAH in a St. line. I. 14. Now •.• z DBG= L FBA, each being a rt. z , adding to *»«wjh z ^J50, we have lABD^lFBC. Ax. 2. Then in as^BL, FB^^ '.' AB=FB, and BD = BC, kxxA i ABD^ i FBG, ... £,JTiD^^FBC. 1.4. Noy O 5r> is double of aABD, on same base PD and between same || s A L, B^ I. 41. and sq. BG is double •' a .FjSC, on same base FB and be- tween sftuie II s FB, GO.- I. 41. ,-, n Bl:=r-({. AK. Now aq. on £0*Bum of O BL and O OL, Mi sum of sq. BO and sq. ^^, ■■sum of sqq. on BA and AG. Q. B. S. Ex. 1. Prove that the square, described upon the diagonal of any given square, is equal to twice the given square. Ex. 2. Find a line, the square on which shall be equal to the sum of the squares on three given straight lines. Ex. 3. If one angle of a triangle be equal to the sum of the other two, and one of the sides contiiiniug this angle being divided into four equal parts, the other contains three of those parts ; the remaining side of the triangle contains five such parts. Ex. 4. The triangles ABC, DEF, having the angles AC It. DFE right angles, have also the sides AB, ^(7 equal to DE, DFy each to each ; shew that the triangles are equal in every respect. Note. This Theorem has been already deduced as a Co- rollary from Prop. E, page 43. Ex. 6. Divide a given straight line into two parts, so that the square on one part shall be double of the square on the other. Ex. 6. If from one of the acute angles of a right-angled triangle a line be drawn to the opposite side, the squares on that aide and on the line so drawn are together equal to the sum of the squares on the segment adjacent to the right angle and on the hypotenuse. Ex. 7. In any trinngle, if a line be drawn from the vertex at right angles to tie Itaso, the difference between the squares on the sides is equal to the dilference between the squares on the segments of the base. 11 76 EUCLWS ELEMENTS. [Book L Proposition XLVIII. Theorem. If the sqiiare described upon one of the sides of a triangle be equal to the squares described upon the oilier two sides of it, the angU contadned by those sides is a right angle. Let the sq. on BO, a side of A ABC, he equal to the sum of the sqq. on AB, AC. Then must l BACbe a rt. angle. From pt. A draw AD a. to AC I. 11. Make ^i)=^jB, and join DC Then vAD=AB, .*. sq. on ^D=sq. on AB ; I. 46, Ex. 2. add to each sq. on AC. then sum of sqq. on AD, -40= sum of sqq. on AB, AC. But •.' z DAC is a rt. angle, .'. sq. on DC = sum of sqq. on AD, AG ; I. 47. and, by hypothesis, sq. on jBC=sum of sqq. on -4J5, ^(7 ; .'. sq. on DC=Bq. on BC ; .'.DO=BU. I. 46, Ex. 3. Then in LB ABC, ADC, -: AB=AP. and ACU common, and BC—DC, :. iBAC=lDAC', La and z DAC is a rt. angle, by construction j .'. z BAC is a rt. angle. Q. & C BOOK 11. INTRODUCTORY REMARKS. The geometrical figure with which we are chiefly concerned in this book is the Rectangle. A rectangle is said to be corv tained by any two of its adjacent sides. Thus if ABCD be a rectangle, it is said to be contained "by AB, ADf or by any other pair of adjacent sides. We shall use the abbreviation rect. AB, AD to express the worrls " the rectangle contained by AB, AD." We shall make frequent use of a Theorem (employed, but not demonstrated, by Euclid) which may be thus stated and proved . Proposition A. Theorem. If the adjacent sides of one rectangle be equal to the adjacent sides of another rectangle, each to each, the rectangles are equal in area. Let ABCD, EFGH be two rectangles : and let AB=EF and BG=FG. Then must rect. ABGD=rect. EFGH. For if the rect. EFGH be applied to the rect. ABCD, to that EF coincides with AB, - then FG will fall on BC, v l EFG= l ABC, and G will coincide with C, V BC=FG. Similarly it may be shewn that H will coincide with D, .». rect. EFGH coincides with and is therefore equal to rect 4JBCD. Q- B. o. 78 EUCLID'S ELEMENTS. [Book if. Be Proposition I. Theorem. If there he two straight lines, one of which is divided into any unriiler of parts, the rectangle contained bif the two straight lines is equal to the rectangles contained by the undivided line and the several parts of the divided line. Let AB and CD be two given st. lines, and let CD he divided into any parts in E, F. Then must rect. AB, CD=sum of red. AB, CE and rect. AB, EF and rect. AB, FD. From C draw CG ± to CD, and in CG make CH=^AB. Through H draw HM \\ to CD. I. 3 1 . Through E, F, and D draw EK, FL, DM \\ to CH. Then EK and FL, being each^C/f, are each = ^J5. Now CM=mn\ of CK and EL and FM. And CM=vcct. AB, CD, :- CH=AB, CK=vect. AB, CE, v CH=AB, EL=vect. AB., EF, '.' EK=AB, FM^xGci. AB,FD, vFL=AB; .: rect. AB, rD = suru of rect. AB, CE and rect. AB, EF and rect. AB, FD. Q. E. D. Ex. If two straight lines be each divided into any number of parts, tlie rectangle contained by the two lines is equal to the rectangles contained by all the parts of the one taken separately with all the parts of the other. cc Book II.] PROPOSITION II. 79 % Proposition II. Theorem. Ij a straight line be (Uvidcd into any two imrts^ the rectangles contained by the 'whole a,nl (dcli of the ijarts are together equai io the sqiujt/re on the ivhole Hue. II •rift 'A m Let the si line AB he divided into any two parts in CL Th en must ;//. on At- ■ sum >J red. AB, AC and rat. AB, CR On A.B describe the sq. ADEB Through Odraw CF |i to AD. Then ^^'=suiii «.,f AF and CK Now AE is the s(j. on AB, AF ■n ct. AB^AC, CJJ=--vvct AB, CB, AD==AB, BE^AB, 1.46. 1.31. sq, ((II ^lii=biiin . f rect. AB, AC and rect. AB, CB, q. E. D. Ex. The square on a straight line is equal to four times the square on half the line. So EUCLID'S ELEMENTS. [Book IL Boo Proposition III. Theorem. If a straight line be divided into any two parts, the rectangle contained bij the ti'hole and one of the parts is equal to the rect ■ angle contained by the two pix/rts together mth ikt sqm/re on tht aforesaid pai-t on\ f & B , I- J > JS Let the st. line AB be divided into any two parts in 0. Then must rect. AB, CB=8um of rect. AC, CB and sq. on CB. Oh CB describe the sq. CDEB. I. 46 From A draw AF \\ to CD, meeting ED produced in F. Then ^ JB;==8um of AD and CR Now AE=Tect. AB, CB, '.' BE^GB, AD=rect. AC, CB, .• CD^CB, CE=^sq. on CB. .'. rect. AB, CB=3um of rect. AC, CB and sq. on CB. Q. E. D. Note. When a straight line is cut in a point, the distance.^) of the point of section from the ends of the line are called the* segments of the line. If a line AB be divided in C, AC and CB are called the internal segments of AR If a line AC he produced to B, AB and CB are called the external segments of AC. Book II.] PROPOSITION IV. 8i ^l Proposition IV. Theorem. If a straight line he divided into any two parts, the iqua/re on the- whole line is e(pial to the squares on the two parts together w^i^h iivice the rectangle contained by the parts. A cy 3 £ F \ i. S a 1 Let the st. line AB be divided into any two parts in G. Then must *q. on AB=sum of sqq. on AC, CB and twice rect. AG, GB. On AB describe the sq. A DEB. I. 46. From AD cut off AH = GB. Then HD=AG. tlvf ^ Draw G(^ II to AD, and HK\\ to AB, meeting CG in F. i 'i ) Then-.-J5Ji:=JiZ, J..:BK=GB, Ax. i. .-. BK, KF, FG, GB are all equal ; and KBG is a rt. z ; .-. GK is the sq. on GB. Def. xxx. Also /fGf=sq. on AG, v HF and HD each^AG. -^oy^ AE^sum oi HG, GK, AF, FF, AE=sq. on AB, and H6/=sq. on AG, CB:=8q. on GB, AF=rect. AG, CB, FE=rect. AG, GB, CF^CB, FG=AGa,ndFK=CB. ,'. sq. on ^jB=sum of sqq. on AG, GB and twice rect. AC, CB. < ' Q. E. D. Ex. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base. Shew that the rectangle, coiitniTiec^ by the segments of the base, is equal to the square on the perpendicular. I ::l ft Sz EUCLID' S ELEMENTS. [Book 11. Proposition V. Theorem. If a straight line he divided into two equal ]>a>is and also into two unequal parts, the rectanyle contaiiied by the unequal parts, to^'^'^her u'ith the scpiare on the line between the points of section, is equal to the square on half the line. 7 ). JB K JL M F Let the st. line AB be divided equally in G and unequally mD. Then must rect. AD, DB together with sq. 07i CD=sq. on CB. tlie ''B. On CB describe Draw DG \\ to CE, uad from it cut off DH=DB. ]Jraw HLK \\ to AD, and AK |1 to DJd. I. 46. I. 31. I. 31. Then rect. Di^=rect. AL, v BF= A C, iind BD=CL. Also LG = sq. on CD, v LH= CD, and HG = CD. Then rect. AD, DB together with sq. on CD = ^jFZ together with if? ' =8umo'^iand CH dnd LG = sum cf DF and CH and LO ■ =aF : ■=sq. on CB, . - ' Q. E. D. rm o\ i\ ih \ II. Book XL] PROPOSITION VI. 83 m ilso of Proposition VI. Tiieokem. If a straight line he bisected and produced to any point, tJie rectangle contained by the lohole line thus prod.ueed and the j art of it produredy toyether with the square on half the line bisected, ■is equal to the square on the straight line which is made up of tite half and the part produced. C a £ -2? II E m Q F Let the st. line AB be bisected in C and produced to P. Then must mi. AD, DB together with sq. on CB=sq. on CD. On CD descriljc the sq. CRFD. 1.46. Draw BG to C£J, and cut off BJf=--BD. L31 Through H draw KLM \ to AD 1.31. Through A draw AK to CE. 2:V Now '.• BG = CD and BH=BD ; • .•.HG=CB', Ax. 3. .'.red. MG = rect. AL. IT. A. Then rect. AD, DB to J -S ^ r/ JB' Let AC^ CD be two st. lines, of which J C is the greater, and let them be placed so as to form one st. line AD. Produce AD to B, making CB=AG. Then -4D=the sum of the lines AC, CD, and D-B=the dift'erence of the lines -(4(7, CD. Theft mu3t difference between sqq. on AC, CD = reel. AD, DB. On OB describe the sq. CEFB. 1. 46. Draw DQ \\ to CE, and from it cut off DH=DB. I. 31. Draw HLK \\ to AD, and AK \\ to DH I. 31. Then rect. DJ^=rect. AL, .'BF^AC, and BD= CL. Also LG = sq. on CD, '.' LH= CD, and IIG = CD. Then dfference between sqq. oii -d(7, CD = difference between sqq. on 05, OZ* = sum of (7/7 and DJ?" =sum of CH and AL =AH =»rect. AD,DH =Tect.AD,DB. Q. E. D. Ex. Shew that Propositions V. and VI. might be deduced from this Proposition. ■Warts are equal to twice the rectangle contained by the ivhole and that part together with the square on the other part. ■4 C Let AB be divided into any two parta in G. Thim must , aqq. on AB, 30= twice red. AB, BC together with sq. on AC. OnAB describe the sq. A DEB. I. 46. From 4Z>cutoff^J/=(75. 1 ^ Draw CF \\ to AD and HGK \\ to AB. I. 31. Then //i?'=sq. on AC, and Oiir=sq. on CB. Hhenaqii. on AB, BC=mm of AE and CK =sum of AK, HF, GE and CK ^mm oi AK, HF Qxidi CE. Hiow AK='rect. AB, BG, :- BK=BC ; CE=Tect. AB, BC, '.' BE^AB -, HF^Hq. on AC. .*. sqq. on AB, jBC= twice rect. AB, 50 together with sq. on AG. Q. E. D, Ex. If straight lines be drawn from G to B and from to p. shew that BGD is a straight line. (■in i 86 EUCLWS ELEMENTS. plook 11. BoJ Proposition VIII. Tueokem. If a straight line be divided into any tim jiartSy four times the rectangle contained bij the vholt; line < half a rt. z , and z ^'(?i'' is a rt. .-. z £;/''(^ishalfart. z ; .-. z EFG= L GEF, and .-. EG=GF. So also z iJi^D is half a rt. z , and BD=DF. , I. 3fi. I. A. I. B. Cor. Now sum of sqq. on A D, DB =sq. on ^i> together with sq. on i)i'' =sq. on ^i'^ I. 47. =sq. on AE together with sq. on EF I. 47. =sqq. on AG, EG together with sqq. on EG, GF I. 47 ■l; = twice sq. on AC together with twice sq. on GF i s» twice sq. on AC together with twice sq. on CD. Q. E. 1 IMAGE EVALUATION TEST TARGET (MT-3) A, 1.0 I.I 1.25 liilM |2.5 ■50 ■"^~ H^H 1.8 1.4 IIJiSi 1.6 ^«> %'^- '^ Photographic Sciences Corporation 23 WIST MAIN STREET WEBSTER, N.Y. 14580 (716)«7a-4503 \ •^ Pf\ V ^8 EVCLms ELEMENTS. [Book n. Proposition X. Theorem. If a straight line be bisected and produced to any point, the square on the whole line thus produced and the square on the part of it produced are together double of the square on half the line bisected and of the square on the liiie made vp of the half and the part produced. ^:] Let the st. line AB be bisected in C and produced to D. Then must mmof sqq. on AD, BD^ twice sum of sqq. on AC, CD. Draw CEa. to AB, and make CE=AC. Join EA, EB and draw EF4 to AD and DF || to CE. Then •.* z s FEB, EFD are together less than two rt. l s, .*. EB and FD will meet if produced towards J5, D in some pt. Q. Join AQ. Then '.• i ACE is a rt. z , .*. L 8 EAC, AEG together=a rt. z , and ••• lEAC= l AEG, .'. z^^C= half art. z. So also z 8 BEG, EBG each = half a rt. z . .•. z AEB is a rt. z . Also z DBG, which =» z EB(}, is half a rt. z , and .'. z BGD is half a rt. z ; .-. ^D=DG. La. 1(1^') I. B. Oor. Again, •.* z FGE=ha\( a rt. z , and z ^^i^G' is art. z , I. 34. .-. z i^EG^= half a rt. z , and EF^FG. I. b. Cor. Then sum of sqq. on AD, DB =sum of sqq. on AD, DG =sq. on AG I. 47. =sq. on AE together with sq. on EG I. 47. =sqq. on AC, EC together with sqq. on EF, FG I. 47. = twice sq. on AC together with twice sq. on EF =■ twice sq. on -40 together with twice sq. on CD. q. b. d.. !f. BooktL] PJfOfOS/TlOAt XI. 89 hit, \on Proposition XI. Problem. To divide a given straight line into two parts, so that the rect- "mgle contained hy the whole and one of the parts shall he equal to the iqua/re on the other part. F (J- given . ADCB. Bisect AD in E and join EB. Produce DA to JP', making EF=BB. On AF descr. the sq. AFGH. Then AB is divided in H so that red. AB, BH=ri Produce GH to K. Then *.• DA is bisected in E and produced to Ff .'. rect. DF, FA together with sq. on AE =sq. on EF -=sq. on EB, '.' EB^EFy B=sum of sqq. on AB, AE. Take from each the square on AE. Then rect. DF, FA=Bq. on AB. 1.46. I. 10. 1.46. sq. on AH. II. 6. 1.47. Ax. 3. Now FK'^rect. DF, FA, v FG=FA. .: FK=AC. Take from each the common part AK. Then FH=^HC', that is, sq. on ^H=rect. AB, BH, '.' BC=AB. Thus AB is divided in H as was reqd. Q. E. F. Ex. Shew that the squares on the whole line and one of the parts are equal to three times the square on the other part, 90 RVCUD'S ELEMENTS. [dookit Boo Proposttion XII. Theorem. In obtuse-angled triangles, if a perpendicular he drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side, upon which, when produced, the perpendi- cula/r falls, and the straight line intercepted ivithout the triangle between the perpendicular and the obtuse angle. Let ABC be an obtuse-angled A, having z AGB obtuse. From A draw AD ± to BG produced. l^hen must sq. on AB be greater than sum of sqq. on BC, CA by twice rect. BC, CD. For since BD is divided into two parts in C, sq. on £I>=sum of sqq. on BC, CD, and twice rect. BC, CD. II. 4. Add to each sq. on DA : then sum of sqq. on BD, D^=8um of sqq. on BC, CD, DA and twice rect. BC, CD. Now sqq. on BD, DA =sq. on AB, I. 47. and sqq. on CD, DA— sq. on CA ; I. 47. .'. sq. on AB=B\im. of sqq. on BC, CA and twice rect. J50, CD. .'. sq. on AB is greater than sum of sqq. on BCf CA by twice rect. BC, CD. Q. E. D. Ex. The squares on the diagonals of a trapezium are together equal to the squares on its two sides, which are not parallel, and twice the rectangle contained by the si'' ^% which are parallel. the the an fa kit Book n.] PROPOSITION XIII. 91 Proposition XIII. Theorem. In every triangle, the square on the side subtending any of the acute angles is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle. Fio. 1. Fio. 2. C JJ C Let ABC be any A , having the z ABC acute. From A draw AD ± to BC or BC produced. Then mud sq. on AC be less than the sum of sqq. on AB, HC, by twice rect. BC, BD. For in Fig. 1 BC is divided into two parts in D, and in Fig. 2 BD is divided into two parts in C ; .'. in both cases sura of sqq. on BC, BD=8\mi of twice rect. BC, BD and sq. on CD. II. 7. Add to each the sq. on DA, then sum of sqq. on BC, BD, DJ.=sum of twice rect. BC, BD and sqq. on CD, DA ; .*. sum of sqq. on BC, -4L'=sum of twice rect. BC, BD and eq. on -40; I. 47. .*. sq. on Ada leas than sum of sqq. on AB, BC by twice rect. BC, BD. The case, in which the perpendicular AD coincides with AC, needs no proof. Q. E. D. Jplx. i-'rove that the sum of the squares on any two sides of a triangle is equal to twice the sum of the squares on half the base and on the line joining the vertical angle with th^ inidclW point of the bas?, itl 'I il EUCLIEfS RLEMBNTS. [Book n. Mook Proposition XIV. Problem. To describe a square that shall be equal to a givm rectilinear figure. Let A be the given rectil. figure. It is reqd. to describe a squa/re that shall^A. Describe the rectangular O BCDE=A. I. 46. Then if BE=ED tlie £7 BGDE is a square, and what was reqd. is done. ■ But if BE be not=^D, produce BE to F, so that EF=ED. Bisect BF in G ; and with centre G and distance C?JB, describe the semicircle BHF. Produce DE to H and join GH. Then, *.• BF is divided equally in G and unequally in E, .'. rect. BE, EF together with sq. on GE =sq. on GF IL 6. =sq. on GH =sum of sqq. on EH, GE. I. 47. Take from each the square on GE. Then rect. BE, EF=aq. on EH. But rect. BE, EF= BD, v EF= ED ; .'. sq. on EH=BD; .'. sq. on ^^= rectil. figure A. (^ B. T» n. nook a.] MISCELLANEOUS EXERCISES. 9> imr MuixllantoMM Exermea on Book IL 1. In a triaBgle, whose vertical angle is a rigDi angle, a straight line is drawn from the vertex perpendicular to the base ; shew that the square on either of the sides adjacent to the right angle is equal to the rectangle contained by the base and the segment of it adjacent to that side. 2. The squares on the diagonals of a parallelogram are to- gether equal to the squares on the four sides. 3. If ABCD be any rectangle, and any point either within or without the rectangle, shew that the sum of the squares on OA^ 00 is equal to the sum of the squares on OB, OD. 4. If either diagonal of a parallelogram he equal to one of the sides about the opposite angle of the figure, the square on it shall be less than the square on the other diameter, by twice the square on the other side about that opposite angle. 6. Produce a given straight line AB to C7, so that the rect- angle, contained by the sum and difference oiAB and J.6',may ne equal to a given square. 6. Shew that the sum of the squares on the diagonals of any quadrilateral is less than the sum of the squares on the four sides, by four times the square on the line joining the middle points of the diagonals. 7. If the square on the perpendicular from the vertex of a triangle is equal to the rectangle, contained by the segments of the base, the vertical angle is a right an^le. 8. If two straight lines be given, shew how to produce one of them so that the rectangle contained by it and the produced part may be equal to the square on the other. 9. If a straight line be divided into three parts, the sqn.ue on the whole line is equal to the sum of the squares on the parts together with twice the rectangle contained by each two of th9 part^. Iliil 94 EUCLID'S ELEMENTS, [Book n. «ooki 10. In any quadrilateral the squares on the diagonals are together equal to twice the sum of the squares on the straight lines joining the middle points of opposite sides. 11. If straight lines be drawn from each angle of a triangle to bisect the opposite sides, four times the sum of the squares on these lines is equal to three times the sum of the squares on the sides of the triangle. 12. CD is drawn perpendicular to AB^ a side of the triangle ABGy in which AC~AB. Shew that the square on CD is equal to the square on BD together with twice the rectangle AD, DB. 13. The hypotenuse JJ5of a right-angled triangle ABC'is trisected in the points D, E', prove that if CD, CE be joined, the sum of the squ.ires on the si if X or. Next suppose we wish to measure two straight lines AB, CD by the same standard EF. If AB contain EF m times and CD ra times, where m and n stand for numbers, whole or fractional, we say that AB and CD are comvunsurable. But it may happen that we may be able to find a standard line EF, such that it is contiiined an exact number of times in AB ; and yet there is no number, whole or fractional, which will express the number of times EF is contained in CD. In such a case, where no unit-line can be found, such that it is contained an exact number of times in each of two lines AB, CD, these two lines are called incommensurable. In the processes of Geometry we constantly meet with incommensurable magnitudes. Thus the side and diagonal of a square are incommensurables ; and so are the diameter and oirciimference of a circle. 96 EUCLID'S ELE^fENTS. [Books 1. k It Next, suppose two lines ABy AG \.o be at right angles to each other and to be commensurable, so that AB contains four times a certain unit of linear measurement, which is contained by -40 three times. A C \ I \ J? Divide ABy AC into four and three equal parts respectively, and draw lines through the points of division parallel to AC^ AB respectively ; then the rectangle ACDB is divided into a number of equal squares, each constructed on a line equal to the unit of linear measurement. If one of these squares be taken as the unit of area, the measure of the area of the rectangle ACDB will be the number of these squares. Now this number will evidently be the same as that obtained by multiplying the measure of AB by the measure of AC; that is, the measure of AB being 4 and the measure of ACS, the measure of ACDB is 4 X 3 or 12. (Algebra, Art. 38.) And generally, if the measures of two adjacent sides of a rectangle, supposed to be commensurable, be a and h, then the measure of the rectangle will be ab. (Algebra, Art. 39.) If all lines were commensurable, then, whatever might be the length of two adjacent sides of a rectangle, we might select the unit of length, so that the measures of the two sides should be whole numbers ; and then we might apply the processes of Algebra to establish many Propositions in Geometry by simpler methods than those adopted by Euclid. Take, for example, the theorem in Book ii. Prop. iv. If all lines were commensurable we might proceed thus — Let the measure of J. be .r, of CJ? ... y, Then the measure of AB is x-^y. Now (x + yf = x^ + y^ + 2x1/, which proves the theorem. lictl fiooks I. k v.] OV Trr StRAStfkEAfES^T OF AkEAS. 97 r>nt. inasmtich as nil lines uie not cotniiicnsurable, we have in Geometry to treat of magnitudea and not of measures: that is, when we use the symbol A to represent a line (as in [. 22), A stands for the line itself and not, as in Algebra, for the number of units of length contained by the line. The method, adopted by Euclid in Book II. to explain the relations between the rectanfjles contained by certain lines, is more exact than any method founded upon Algebraical prin- ciples can be ; because his method applies not merely to the case in which the sides of a rectangle are commensurable, but also to the case in which they are incommensurable. The student is now in a position to understand the practical application of the theory of Equivalence of Areas, of which the foundation is the 35th Proposition of Book I. We shall give a few examples of the use made of this theory in Men- suration. ..: . Area of a Parallelogram. The area of a parallelogram ABCD is equal to the area of the rectangle ABEF on the same base AB and between the same parallels ^jB, i^C Now BB is the altitude of the parallelogram ABCD if 4B be taken as the base. Hence area of O ^£Ci)=rect. AB, BE. If then the measure of the base be denoted by &, and altitude h, the measure of the area of the O will be denoted by hK That is, when the base and altitude are commensurable, measure of area = measure of base into measure of altitude. ^ EVCUDS ELEMENTS. [Books 1. k It Bo Arm of a Triangle. If from one of the angular points A of a triangle ABC, a perpendicular AD be drawn to BC, Fig. 1, or to BC produced, Fig. 2, Fio. 1. C and if, in both enses, a parallelogram ABCE be completed of which AB, BC are adjacent sides, area of A ^50= half of area of O ABCE. Now if the measure of BC be 6, and AD...h, measure of area oiCJ ABCE is ftfc ; hh .*, measure of area of A ABC is 2 Area of a Rhombus. Let ABCD be the wiven vhoTubus. Draw the diagonals ^Oand BD, cutting on^ another in 0. It is easy to prove that A C and BD bisect each other at right angles. Then if the measure of ^0 be x, and BD ... i/, measure of area of rhombus = twice measure of A ACD. = twice ^ 4 I. k If. BookiLftllJ A^!/:;A 01' A TRAPLZtUM. 99 luced, Ana of a Trapexium. Let ABCD be the given trapezium, having the sides AB, Ci) parallel. Draw AE a.t right angles to CD. I [0 I Produce DO to J", making aF= ^5. Join ^jP, cutting £0 in 0. Then in /^b AOB, COF, '.' iBAO= L CFO, and z AOB=> L FOC, and AB=CF; .\ lCOF=^ t^AOB. 1.26. Hence trapezium ABCD= a ADF. Now suppose the measures of AB^ CD, At] to he m, w, p respectively ; .'. measure of DJ?'=w + n, •.* CF=AB. Then measure ot area of trapezium = i (measure of DF X measure of AE) =^(m + w) Xp. That is, the measure of the area of a trapezium is fouk (,; multiplying half the measure of the sum of the parallel ijides by the measure of the perpendicuHr distappp bptwepn tjjf) parallel sides, ■"I - ' ■■ -^■" ■ lOO EUCLIiys ELEMENTS. [Books I. ft II. Area of an Irregular Polygon. There are three methods of finding the area of an irregular polygon, which we shall here briefly notice. I. The polygon may he divided into triangles, and the area of each of these triangles be found stparately. J? V D Thus the area of the irregular polygon ABODE is equal to the sum of the areas of the triangles ABE, EBD, DBG. II. The polygon may he co. verted into a single triangle of equal area. If ABODE be a pentagon, we can convert it into an equivalent quadrilateral by the following process • js — i> Jf Join BD and draw OF parallel to BD, meeting ED pro- duced in F, and join BF. Then will quadrilateral ABFE=-penti\^o\\ ABODE. For A BDF= lBOD, on same base BD and between same parallels. If, then, from the pentagon we remove lBOD, and add A BDF to the remainder, we obtain a quadrilateral ABFB equivalent to the pentagon A BOD^. fiooks t. k II.] AJ?£A OF AN IRREGULAR POLYGON. lOl The quadrilateral may then, by a similar process, be con- verted into an equivalent triangle, and thus a polygon of any number of sides may be gradually converted into an equiva- lent triangle. » The area of this triangle may then be found. III. The third method is chiefly employed in practice by Surveyors v^ / c T\ \ \ N / / a Let ABCDEFG be an irregular polygon. Draw AE, the longest diagonal, and drop perpendiculars oil AE from the other angular points of the polygon. The polygon is thus divided into figures which are either right-angled triangles, rectangles, or trapeziums ; and the areas of each of these figures may be readily calculated. 102 MVCLID'S ELEMENTS, [Bodka I. k II. Note 7. On Projections. The projection of a point B, on a straight line of unlimited length AE, is the point M at the foot of the perpendicular dropped from B on AE. The projection of a straight line BG, on a straight line of unlimited length AE, is MN, — the part of AE intercepted between perpendiculars drawn from B and C. When two lines, as AB and AE, form an angle, the pro» jection of AB on AE is AM. We might employ the term projection with advantage to shorten and make clearer the enunciations of Props, xii. and XIII. of Book II. Thus the enunciation of Prop. xii. might be : — " In oblique-angled triangles, the square on the side sub- tending the obtuse angle is greater thiin the squares on the sides containing that angle, by twice the rectangle contained by one of these sides and the pTojection of the other on it." The enunciation of Prop. xiii. might be altered in a similar manner. I. ii II. Books L ft II.] ON" LOCI. 103 Note 8. On Lod. Suppose we have to determine the position of a point, which is equidistant from the extremities of a given straight lineBa There is an infinite number of points satisfying this con- dition, for the vertex of any isosceles triangle, described on BG as its base, is equidistant from £ and Oc m Let ABC be one of the isosceles triangles described on BC. If BC be bisected in I), MN, a perpendicular to BC drawn through D, will pass through A. It is easy to shew that any point in MN, or MN produced in either direction, is equidistant from B and C. It may also be proved that no point out of MN is equi distant from B and C. The line MN is called the Locus of all the points, infinite in number, which are equidistant from B and G. Dbf. In plane Geometry Locus is the name given to a line, straight or curved, all of whose points satitfy a certain geometrical condition (or have a common property), to tlif exclusion ,of all other points. 104 EUCLID'S ELEMENTS. [Books I. & IL Next, sujipose we have to determine the position of a point, which is equidistant from three given points A^ B, C, not in the same straight line. If we join A and B, we know that all points equidistant from A and B lie in the line FD, which bisects ^^ at right angles. If we join B and C, we know that all points equidistant from B and C lie in the line QE, which bisects BC at right angles. Hence 0, the point of intersection of PD and QE^ is the only point equidistant from A , B and 0. PD is the Locus of points equidistant from A and B, QE -Band 0, and the Intersection of these Loci determines the point, which is equidistant from A, B and C. Examples of Loci. Find the loci of (1) Points at a given distance from a given point. (2) Points at a given distance from a given straight line. (3) The middle points of straight lines drawn from a U'i"en point to a given straight line. (4) Points equidistant from the arms of an angle. (5) Points equidistant from a given circle. (6) Points equally distant from two straight lines whicb intersect. ft II. point, not in Books I. ft II.] SOLUTION OF PROBLEMS. 105 Note 9. On. the Methods employed in the solution of Problems. In the solution of Geometrical Exercisos, certain methods maybe applied with success to particul;i classes of questions. We propose to make a few remarks on these methods, so far as they are applicable to the first two books of Euclid's Elements. The Method of Synthesis. In the Exercises, attached to the Propositions in the pre- ceding pages, the construction of the diagram, necessary for the solution of each question, has usually been fully described, or sufficiently suggested. The student has in most cases been required simply to apply the geometrical fact, proved in the Proposition preceding the exercise, in order to arrive at the conclusion demanded in the question. This way of proceeding is called Synthesis (o-wi/tfeo-trs= com- position), because in it we proceed by a regular chain of reason- ing from what is given to what is sought. This being the method employed by Euclid throughout the Elements, we have no need to exemplify it here. The Method of Analysis, The solution of many Problems is rendered more easy by supposing the problem solved and the diagram constructed. It is then often possible to observe relations between lines, angles and figures in the diagram, which are suggestive of the steps by which the necessary construction might have been effected. This is called the Method of Analysis (ai/aXvo-tj™ resolution). It is a method of discovering truth by reasoning concerning things unknown or propositions merely supposed, as if the one were given or the other were really true. The process can best be explained by the following examples. Our first example of the Analytical process shall be Uie 31st Proposition of JilugliU's First fJook, iM •1 io6 EUCLID'S ELEMENTS. [Books J. ft I! Ex, 1. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC be the given straight line. Suppose the problem to be effected, and UF to be the straight line required. E A J' B n ^ Now we know that any straight live AD drawn from A to meet BG makes equal angles with EF and BC. (i. 29.) This is a fact from which we can work backward, and arrive at the steps necessary for the solution of the problem ; thus : fake any point I) in i>'0, join AD, make l EAD= l ADC, and produce EA to F: then EF must be parallel to BC. Ex. 2. To inscribe in a triangle a rhombus, having one of its angles coincident with an angle of the triangle. Let ABC be the given triangle. Suppose the problem to be effected, and DBFE to be tht rhombus. Then if EB be joined, i DBE= l FBE. This is a fact from which we can work backward, and deduce the necessary construction ; thus : Bisect z ABC by the straight line BE, meeting AC in E. Draw ED and EE paraUel to BC and AB respectively. Then D^FE is the rhoijibus required. (See Ex. 4, p. 5c'.) fiooks I. & II J SOLCriOAt OF PROBLEMS. 167 Ex. ;}. To determine the foint in a given straight line, at which straight lliie.% drawn from two givin ]mnts, on the same side of the given line, make equal angles with it. Let CD be the given line, and A and B the given points. Suppose the problem to be effected, and P to be the point required. D We then reason thus : • If BP were produced to some point A, L CPA\ being= z BPD, will be= z APC, Again, if PA' be made equal to PA, A A' will be bisected by GP at right angles. This is a fact from which we can work backward, and find the steps necessary for the solution of the problem ; thus : From A draw AO ± to CD. " Produce -40 to A', making 0-4'= 0-4. Join BA', cutting CD in P. / Then P is the point required. 'I i Note 10. On Symmetry. The problem, which we have just been considering, suggests the following remarks : If two points, A and -4', be so situated with respect to a straight line CD, that CD bisects at right angles the straight line joining A and A', then A and A' are said to he symmetrical with regard to CD. The importance of symmetrical relations, as suggestive of methods for the solution of problems, cannot be fully shewn io8 EUCUIfS: ELEMENTS. [Books l.hVL to a learner, who is unacquainted with the properties of the circle. The following example, however, will illustrate this part of the subject sufficiently for our purpose at present. ¥in^ a point in a given straight line, such that the sum of its distances from two fixed points on the same side of the line is a minimum, that is, less than the sum of the distances of any other point In the line from the fixed points. Taking the diagram of the last example, suppose CD to be the given line, and A, B the given points. Now if A and A' be symmetrical with respect to CD, we know that every point in CD is equally distant from A and A'. (See Note 8, p. 103.) Hence the sum of the dbtances of any point in CD from A and B is equal to the sum of the distances of that point from A'andB, But the sum of the distances of a point in CD from A' and B is the least possible when it lies in the straight line joining A' and B. Hence the point P, determined as in the last example, is the point required. NoTB. Propositions ix., x., xi., xii. of Book I. give good examples of symmetrical constructions. NotbU. JEucUd^s Proof of T, i. The angles at the base of an isosceles triangle are equal to one another ; and if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an isosceles A , having AB^a 40 Produce AB, AC to D and E, Then must L ABC= l ACB, z ^GB. 1.4. Ax. 3. 1.4. Ax. 8. Q. E. D. ltd flVcr. Ib'S MlkMEN TS. [ftook» t ik It. h. Note 12. EiicMs Proof of I. 6. If two anglet of a triangle be equal to one another^ the sides also, whiek tuhtend the egucU angles, shail be equal to one another. In L ABO let iACB^ lABa Then must AB= AC. For if not, ABw either greater or lesa than AC Suppose ^5 to be frrenter than AC From AB cut off BD=AG, and join LC, Thenin A a DBG, AC B, * •.• DB^'AG, and BC is coinnion, and i DBC= L AGB, ,\ lDBG^lACB; I. 4. that is, the less = the greater ; which is absurd. > .*. AB is not greater than AG. Similarly it may be shewn that AB is not less than AC\ .'.AB'^AG. , ' ■ Q. E. D. Note 13. EuclicPs Proof of I. 7. Upon the same base and on the same side of it, there cannot be two triangles that have their sides which are ter- minated in one extremity of the base equal to one another, and their sides which are terminated in the other extremity of the base equal also. If it be possible, on the same base AB, and on the same side of it, let there be two iisAGB, ADBf such that ^C» J A and also BC=5I>. Join CD. Books I. ft A.] EUCLID'S PROOF OF I. ^. III Firsts wheu the vertex of each of the Zis is outside the other A (Fig. 1.) ; Fio 1. Pio. 2. 1.5. .-. lACD^lADO. But lACDv& ^eater than z BCD ; . . I ADC is greater than l BCD ; much more is i BDC greater than z BCD, Again, •/ BC=BD, .'. L BDC-= L BCD, that is, z BDC is both equal to and greater than z BCD ; ivhich is absurd. Secondly, when the vertex D of one of the As falls within the other a (Fig. 2) ; Produce ^ C and ^D to ^ and F Then '.• AC=AD. .: L ECD= L FDC. I. 6. But z BCD is greater than z BCD ; .-. z FDC is greater than z ifOD ; much more is z JBI>0 greater than z BOD. Again, ••• BC=BD, .-. z BD(7= z JBCP ; that is, z -BDO is both equal to and greater than z BCD ; which is absurd. Lastly, when the vertex D of one of the as falls on a side BC of the other, it is plain that BO and BD cannot benquul, <«•=•». 112 EUCLID'S ELEMENTS. [Books L & IL Note 14. Euclidi Proof of I. 8. If tivo triangles have two aides of the one equal to two sides of the otJier, each to each, and have likemse their bases equal, the angle which is contained hy the two sides of the one must bt equal to the angle contained by the two aides of the other. Let the sides of the As A} ', DBF be equal, each to each, that is, AB^DE, AG=^DF and BC'-EF. Then must iBAC=' L EDF, Apply the a ^50 to the aDEF. so that pt. B is on pt. E, and BC on EF, Then •/ BC^EF, .'. C will coincide with F, and BC will coincide with EF. Then -4 B and AC must coincide with DE and DF. For if AB and AG have a different position, as OE, GF, then upon the same base and upon the same side of it there can be two as, which have their sides which are terminated in one extremity of the base equal, and their sides which are ter- minated in the other extremity of the base also equal : which is impossible. I. 7. .*. since base BC coincides with base EF, AB must coincide with DE, and AG with DF ; .'. I fJC coincides with and is equal to 4. EDF. Q. E, p. Book I. ft II.] ANOTHER rROOF OF /. 24. "3 Note 15. ^no^/((r 7Voo/ 0/ i. 24. In the AS ABC, DIJF, let All=DE und JC'=i>2^,and let L fiAC be },neiiter tluui i K1)F. Tl -11 viuat BC be yreakr than EF. A A Apply the a DEF to the A ABO so that DE coincides with AB. 'fheri ••• I EDF is less than / BAG, DF will fall between BA and AG, . »d F will fall on, or above, or 6eZow, BC. I. If i' fall on BG, BF is less than BG\ .'. EF is less than BG. II. If F fall a&ove i?^, ^i'', FA together are less than BG, GA, and FA = G A; .'. £i^ is less than 50; .'. EF is less than BC. III. UF Ml below BG, let J i?' cut £C in 0. Then BO, OF together are greater than BF, I. 20. and OG. AG AG; I. 20. .'.BG,AF £i^, ^0 together, and AF=AG, .'. BG IS greater than BF • and .'. EF is less than BC, Q. e. d. 8 '1 tl 4. u.. I!4 EUCLID'S ELEMENTS. [Books I. & tt Note 16. Euclid's Proof of I. 26. If two triangles have two angles of the one eqiial to two angles of the other, each to each, and one side equal to one side, viz., either the sides adjacent to the equal angles, or the sides opposite to equal angles in each ; then shall the other sides be equal, each to each ; and also the third angle of the one to the third angle of the other. C J3 In ^s ABC, DEF, Let I ABC = L DEF, and l ACB = i DFE ; and first, Let the sides adjacent to the equal /: s in each be equal, that is, let ^C=^'i*'. Then must AB=DE,and AC=lJF,dnd l BAC = i EDF. For if AB be not=DE, one of them must be the greater. Let AB he the greater, and make GB=DE, and join GC Then in /^fiGBC,DEF., V GB=DE, and BC=EF, and z GBC = / DEF, .-. z GCB= . DFE. I. 4. But z^CB= z DFE by hypothesis; .-. iGCB= lACB; that is, the less =the greater, which is impossible. .'. AB is not greater than DE. In the same way it may be shewn that AB is not less than DE; .'. AB=DE. Then in As ABC, DEF, \- AB= DE, and BC = EF, and / ABC= l DEF, .-. AC=DF, and i BAC= l EDF. I. 4. tcIL Books 1. & 11.1 EVCUD'S rROOF OP' J. 26. I"! Nexty let the sides which are opposite to equal angles in o.ti'.i t' imj^le be equal, viz., AB=DE. Tten must AC=DF, and BC=EF, and l BAG = -1 EDF. Forif JBO be not=^J?', let BO be the greater, and make BH=EF, &nd join AH. Then m AS ABH,DEF, V AB=DE, and BH^EF, and z ABH= l BEF, .-. L AHB= L DFE. I. 4. "BvX lACB=^ L DFE, by hypothesis, .'. z AHB = z ^(75 ; that is, the exterior z of a AHC is equal to the interior and opposite z ACBf which is impossible. .'. BC is not greater than EF. In the same way it may be shewn that BC is not less than FF; .'.BC^EF. f Thenin As^J50, DEi^, •.• AB=DE, an.l BC^EF, and i ABG= L BEF, .'. AC==DF, and i BAC^ l EDF. I. 4. Hi I ■ mi R|: Q. B. D. n6 EUCLID'S ELEMENTS. \tooliB 1. & tt Miscellaneous Exercises on Bodies I. and II, 1. AB and CD are equal straight lines, bisecting one another at right angles. Shew that ACBD is a square. 2. From a point in the side of a parallelogram draw a line dividing the parallelogram into two equal parts. 3. In the triangle FDCj if FCD be a right angle, and angle FDG be double of angle CFD, shew that FD is double of DC. 4. li ABChe an equilateral triangle, and AD, BE be per- pendiculars to the opposite sides intersecting in F ; shew that the square on AB is equal to three times the square on AF. 5. Describe a rhombus, which shall be equal to a given triangle, and have each of its sides equal to one side of the triangle. 6. From a given point, outside a given straight line, draw a line making with the given line an angle equal to a given rectilineal angle. 7. If two straight lines be drawn from two given points to meet in a given straight line, shew that the sum of these lines is the least possible, when they make equal angles with the given line. 8. ABGD is a parallelogram, whose diagonals AG^ BD in- tersect in ; shew that if the parallelograms AOBPy DOGQ be completed, the straight line joining P and Q passes through 0. 9. ABGD, EBGF are two parallelograms on the same base BG, and so situated that CF passes through A. Join DF, and produce it to meet BE produced in K ; join FB, and prove that the triangle FAB equaLi the triangle FEK. 10. The alternate sides of a polygon are produced to meet ; shew that all the angles at their points of intei-section together with four right angles are equal to all the interior angles of the polygon. 11. Shew that tne perimeter of a rectangle is always greater than that of the square equal to the rectangle. Books I. & TI.] MTSCELLANROr^ EXERCISES. "7 12. Shew thiit the opposite sides of an equiangular hexagon are parallel, though they be not equal. 13. If two equal straight lines intersect each other anywhere at right angles, shew that the area of the quadrilateral fornud by joining their extremities is invariable, and equal to one-luill the square on either line. 14. Two triangles ACB, ADB are constructed on the same side of the same base AB. Shew that if AC—BD ami AD=BC, then CD is parallel to AB ; but if AG=BCii\\^\ AD—BD^ then CD is perpendicular to AB. 16. AB is the hypotenuse of a right-angled triangle ABC : find a point D in AB^ such that DB may be equal to the per- pendicular from I> on AG. 16. Find the locus of the vertices of triangles of equal area on the same base, and oti the same side of it. 17. Shew that the. perimeter of an isosceles triangle is less than that of any triangle of equal area on the same base. 18. If each of the equal angles of an isosceles triangle be equal to one-fourth the vertical angle, and from one of them a perpendicular be drawn to the base, meeting the opposite side produced, then will the part produced, the perpendicular, and the remaining side, form an equilateral triangle. 19. If a straight line terminated by the sides of a triangle be bisected, shew that no other line terminated by the same two sides ciin be bisected in the same point. 20. Shew how to bisect a given quadrilateral by a straight line drawn from one of its angles. 21. Gi vcn the lengths of the two diagonals of a rhombus, con- struct it. 22. ABCD is a quadrilateral figure : construct a triangle whose base shall be in the line AB, such that its altitude shall be equal to a given line, and its area equal to that of the quadrilateral. 23. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides, their sum will be equal to the perpendicular from either extremity of the base upon the opposite side, It' m i.'rf. m ii8 EUCLTiys ELEMRATTS. [Books I. & IL 24. If ABC be a triangle, in which C is a right angle, and DE be drawn from a point D in ^0 at right angles to AB, prove that the rectangles AB^ AE and AC, AD are equal 25. A line is drawn bisecting parallelogram ABCD, and meeting AD, BC in E and F : shew that the triangles EBF, CED are equal. 26. Upon the hypotenuse BG and the sides CA, AB of a right-angled triangle ABC, squares BDEC, AF and AG are described : shew that the squares on DG and EF are together equal to five times the square on BC. 27. If from the vertical angle of a triangle three straight lines be drawn, one bisecting the angle, the second bisecting the base, and the third perpendicular to the base, shew that the first lies, both in position and magnitude, between the other two. 28. If ABC be a triangle, whose angle ^4 is a right angle, and BE, CF be drawn bisecting the opposite sides respectively, shew that four times the sum of the squares on BE and CF is equal to five times the square on BC. 29. Let ACB, ADB be two right-angled triangles having a common hypotenuse AB. Join CD and on CD produced both ways draw perpendiculars AE, BF. Shew that the sum of the squares on CE and CF is equal to the sum of the squares on DE and DF. 30. In the base ^C of a triangle take any point D: bisect AD, DC, AB, BC at the points E, F, G, H respectively. Shew that EG is equal and parallel to J'*^. 31. If AD be dra\^n from the vertex of an isosceles triangle ABC to a point D in the base, shew that the rectangle BD, DC is equal to the difference between the squares on AB and AD. 32. If in the sides of a squart four points be taken at equal distances from the four angular points taken in order, the figure contained by the straight lines, which join them, shall also be a square. 33. If the sides of an equilateral and equian;;ular pentagon be produced to meet, shew tliat the sum of the angles at the points of meeting is equal to two right angles, Books 1 k II.] MiSCELL4NE0ZtS EXERCISES. J|0 I 34. Describe a square that shall be equal to the diilerence between two given and unequal si^uares. 35. ABCD, AECF are two parallelograms, EA, AD being in a straight line. Let FG, drawn parallel to AC, meet BA produced in G, Then the triangle ABE equals the triangle AUG. 36. From AC, the diagonal of a square ABCD, cut off AE equal to one-fourth of AC, and join BE, DE. Shew that the figure BADE is ec^ual to twice the square on AE. 37. If ABC be a triangle, with the angles at B and C each double of the angle at A, prove that the square on AB is equal to the square on BC together with the rectangle AB, BG. 38. If two sides of a quadrilateral be parallel, the triangle contained by either of the other sides and the two straigiit lines drawn from its extremities to the middle point of the opposite side is half the quadrilateral. 39. Describe a parallelogram equal to and equiangular with a given parallelogram, and having a given altitude. 40. If the sides of a triangle taken in order be produced to twice their original lengths, ana the outer extremifies be joined, the triangle so formed wijl oe seven times the orisinal triangle. 41. If one of the acute angles of a right-angled isosceles triangle be bisected, the opposite side will be divided by the bisecting line into two parts, such that the square on one will be double of the square on the other. 42. ABC is a triangle, right-angled at B, and BD is drawn perpendicular to the base, and is produced to E until KCB is a right angle ; prove that the square on BC is equal to the sum of the rectangles AD, DC and BD, DE. 43. Shew that the sum of the squares on two unequal lines is greater than twice the rectangle contained by the lines. 44. From a given isosceles triangle cut off a trapezium, having the base of the triangle for one of its parallel sides, and having the other three sides equal. 11 11^: 3 ill W ■ 'M I, i' tM nUCLTD'S ELEMENTS. [Books J ^ It 45. Jf any number of paiallelofjiams be constructed hadng their siilcs of given length, shew that the sum of the squares on the diagonals of each will be the same. 46. ABCD is a right-angled parallelogram, and AB is doub!e of BC ; on AB an equilateral triangle is constructed : shew that its area will be less than that of the parallelogram. 47. A point is taken within a triangle ABC, such that the angles JBOC, CO A, AOB are equal ; prove that the squares on BCf CA, AB are together equal to the rectangles contained by OB, OC; DC, OA ; OA, OB; and twice the sum of the squares on OA, OB, 0(1 48. If the sides of an equilateral and equiangular hexagon be produced to tiioet, the angles formed by these lines are together equal to four right angles. 49. ABC is a triangle right-angled at A ; in the hypottv nuse two points D, E are taken suth that BD = BA and CE=CA ; shew that the square on DE is equal to twice tVe rectangle contained by BE, CD. 50. Given one side of a rectangle which is equal in area to er than, the radius ot the circle. Dbf. I. A straight line, as PQ^ drawn so as to cut a circle ABCDf is called a Secant. r iii J!' That such a line can only meet the circumference in two points may be shewn thus : Some point within the circle is the centre ; let this be 0. Join OA. Then (Ex. 1, i. 16) we can draw one, ana only one, straight line from 0, to meet the straight line* F^, such that it shall be equal to OA, Let this line be 00. Aien A and G are the only points in PQ, which are on tKj circumLirence of the circle. B. B. II. II W Hi Wct.riys elemknts. [Book l!t Def. it. The portion AG of the secant PQt intercepted by the circle, is called a Chord. Def. III. The two portions, into which a chord divides the circumference, as ABC and ADC, are called Arcs. Def. IV. The two figures into wh'ch a chord divides the circle, as ABC and ADC, that is, tlie figures, of which the boun- daries are respectively the arc ABC and the chord AC, and the arc ADC and the chord AC, are called Seg::knts of the circle. Def. V. The figure AOCD, whose boundaries are two radii and the arc intercepted by them, is ciilled n Srctor. Def. VI. A circle is said to be described about a rectilinear figure, when the circumference passes through each of the angular points of the figure. And the figure is said to be inscribed in the circia Book III.] rKOPos/riox /. 123 Proposition I. Theorrm. The U c, which bisects a chord of a circle at right angles f must contain the centre. . Let ABC be the given ©. Let the st. line CE bisect the chord AB at rt. angles in D, Then the centre of the © must lie in CE. For if not, let 0, a pt. out of CE, be the centre ; and join OA, OD, OB. Then, in as ODA, ODB, V AD = BD, and DO is common, and OA = OB ; .: A ODA = I ODB; L c. and .-. I ODB is a right z . L Def. 9 But z GDB is a right z , by construction ; .'. z ODB = z CDB, which is impossible ; .*. is not the centre. Thus It may be shewn that no point, out of CE, can be the centre, and .". the centre must lie in CE. Cor. // the chord CE he bisected in Fj then f is the centra 0/ thf circle, Bo( "4 EUCr.TD'S ELEMENTS. [Book TIL Proposition II. Theorem. If any two foinh he ttdeu in the circumference of a circle, the straight line, which joins therrif must fall within the circle. Let A and B be any two pts. in the Oce of the ® ABC. Then must the st. line AB fall within the . Take any pt. D in the line AB. Find the centre of the © . III. 1, Cor. Join OA, OD, OB. Then-.- iOAB = lOBA, La. and z ODB is greater than z OAB, I. 16. .*. z ODB is greater than z OB A ; and .'. OB is greater than OD. I. 19. .•. the distance of D from is less than the radius of the ©, and .*. D lies within the © . Post. And the same may be shewn of any other pt. in AB. ,'. AB lies entirely within the ®. Book III.] PROPOSITION III. 125 Proposition III. Theorem. If a straight line, dravm through the centre of a circle, bixeH a chord of the circle, which does not pass through the centre, it must cut it at right angles : and conversely, if it cut it at right angles, it mu^t bisect it. In the ABC, let the chord AB, which does not pass through the centre 0, be bisected in E by the diameter CD. Then must CD he ± to AB. Join OA, OB. Then in A s AEO, BEO, V AE=BE, and EO is common, and OA=^OB, .: L OEA^L OEB. Hence OE is ± to AB, that is, CD is ± to AB. Next let CD be ± to AB. Then must CD bisect AB. La I. Def. 9. For I. E. Cor. p. 43. Q. £. D. .* OA = OB, and OE is common, in the right-angled as AEO, BEO, .'. AE=BE, that is, CD bisects AB. Ex. 1. Shew that, if CD does not cut AB&t right angles, it cannot bisect it. Ex. 2. A line, which bisects two parallel chords in a circle, is also perpendicular to them. Ex. 3. Through a given point within a circle, which is not (b9 ^ntre, dr^w a chord which shall 1^ Vi^ect^d jp th)it point, in i "(.I > li 126 EUCLID'S ELEMENTS. [Book III Bo Proposition IV. Theorem. If in a circle Uoo chords, which do not both pans through the centre, cut one another, they do not bisect each other. Let the chords AB, CD, which do not both pass through the centre, cut one another, in the pt. E, in the ® ACBD. Then AB, CD do not bisect each other. If one of them pass through the centre, it is plainly not bisected by the other, which does not pass through the centre. But if neither pass through the centre, let, if it be possible, AE=EB and CE=ED ; find the centre 0, and join OE. Then '.* OE, passing through the centre, bisects AB, .: L OEA is a rt. z . III. 3. And •.• OE, passing through the centre, bisects CD, .-. z OEC is a rt. z ; III. 3 /. z OEA = z OEC, which is impossible ; ,*. AB, CD do not bisect each other. q. e. d . Ex. 1. Shew that the locus of the points of bisection of ; parallel chords of a circle is a straight line. Ex. 2. Shew that no parallelogmm, except those which are rectangular, can be inscribed in a ciivle, Book Itl.] PROtOSlTIO^ V, -r PllOrosiTloN V. TUKttltKM. y/ two circles ciit one another, lluij cannot have the same centre. I. Def. 13. I. Def. 13. If it be possible, let be the common centre of the ®s aBCj ADC, which cut one anotlier in the pts. A and C. Join OA, and draw OEF meeting the ©a in £' and F. Then •.* is the centre of © ABC, .\OE=OA\ and *.• is the centre of © ADC, .'. OF=OA ; .'. 0JE7=01'', which is impossible ; .*. is not the common centre. Q. E. D. Ex. If two cijcli's (lit one another, shew that a line dawn thioujih a point oi inteisi ction, teniniuitcd by tlio cirtuiiil'er- ences and parallel to the line joining' ilie centres, is double of the line joining the centres. Note. Ciicles which have the same centre are called Con^ centric. m K 1 t2d EUCUb'S eLEMeMTs. (Book 111 Note 1. On the Contact of Circles. Def. YII. Circles are said to touch each other, which meet but do not cut each other. One circle is said to touch another internally, when one point of the circumference of the former lies on, and no point iclthout, the circumference of the other. • Hence for internal contact one circle must be smaller than the other. Two circles are said to touch externally, when one point of the circumference of the one lies on, and no point mthin the circumference of the other. N.B. No restriction is placed by these definitions on the number of points of contact, and it is not till we reach Prop. XIII. that we prove that there can be but one point oj contact. 11. Book III.1 pROPosmoN- VI ii^ I ■ I- ■ If Proposition VI. Theorem. If one circle touch another internally, they cannot have the same centre. iK Let AD.^ touch © ABG internally, and let -4 be a point of contact. Then some point E in the Oce ADE lies within © ABG. Def. 7. If it be possible, let be the common centre of the two ® s. Join OA, and draw OEC, meeting the Oces in E and C. Then *.• is the the centre of © ABG, .. OA = OG', I. l>ef 13. and '.• is the centre of © ADE, .-. OA = 0E. I. Def. 13. Hence OE=OQ, which is impossible , ,'. O is not the common centre of the two © s Q. E. D. Iv fid EUCLID'S ELEMENTS. iliook lit Proposition VII. Titeouem. If from any point within a circh., which is not the centre, 4<-aiffht lines he drawn to the circumference, the greatest of these lines is that which passes through the centre. Let ABC he a © , of which is the centre. From F, any pt. within the 0, draw the st. line PA, pass- inji throujjh and nieetinsj tlie Oce in A. Then must PA be greater tha,n any other st. line, drawn from P to the Oce. For let PB lie any other st. line, drawn from P to meet thft Oce in B, and join BO. Then \' AO=BO, .'. ^P=siun of BO and OP. But the sum <«f BO and OP is greater than BP, I. 20. and .•. AP is greater than BP. q. e. d. Ex. 1. If AP be produced to meet the circumference in l>, shew that PD is less than any other straight line that can l,e drawn from P to the circumference. Ex. 2. Shew that PB continually decreases, as B passes trom A to D. Ex. 3. Shew that two straight lines, hut not three, that shall be equal, can be drawn iroiii P to the circuiuierence. Bjtk i:i.] /7CO/VS/770X VIJI. 131 ;, : ,■ PliOl'.),,lil(iN Vlll. ThiOURELI. 7/ from any fohtt ivitl out a circle, straight lines be drawn to the circumference, the l(ci. t of thc^c llihen is that irliidt,, u-hca pro- duced, passes thioiKjh the centre, ail d the yrn}- -A is that which passes through the <:<;, in. Let ABC be a ©, of which is the centre. From P any pt. outside the ©, draw the st. line PA^G, n)eetit)g the Oee in A and C. Then must FA be less, and PC greater, than any other si 'iw^ drawn from P to the Oce. For let PB be any other st. line drawn from P to nieei, the Cce in B, and juin BO. 'i hen •.• sum of PB and BO is jrveater than OP, i. .0. .•. ,suin of PB and BO is greater than sum of AP and jiO. IkitBO=AO; .'. PB is greater than AP. Again *.' PB is less than the sum of PO, OB, .'. PB is less than tlie sum of PO, OC ; .-. PB is less than PC. Ex. 1. Chew that PB continually increases as B passes fruui A to C. Ex. 2. Shew that from P two straight lines, but not tnree, that shall be eoual, can be drawn to the circumference. Note. From Props, vii. and viii. we deduce the follov.ing Corcltiry, whicn v/e shall use in the proof of Props, xi. and xiii. Cor. If c 'point he taken^ luithin or xcithoxit a circle, oj all straight lines draivn from it to the circumference, the greatest is fjjicii ^YtQli Tiicc-i the circumferei, ce after 'passing through the centra, 1.20. Q. E. C. i|S 'tT; ill 132 EUCLID'S ELEMENTS. [Book Til. FuoposTTioN IX, Theorem. Ij a point be taken within a circle, from, which there JclU mort than two equnl strrmfiii lines to the circumference, that point m ihe centre of the circle. Let be a pt. in the © ABC tVoiii which more than two st. lines OA, OB, OG, drawn to the Oce, are equal. Then must he the centre of the (S). • ' Join AB, BC, nud draw OD, OE i. to AB, BG. Then '.• OA = OB, and OD is common, in the right-angled a s ^ OD, BOD, .'. AD=DB; .'. the centre of the © is in DO. Similarly it may be shown tha: the centre of the Iz £10 ; ' /, ii the centre of the © . Q.B, D. I. E. Cor. p. 43. III. 1. fiook in.] PtiOPOSITION X. t33 Proposition X. Thkorem. Two circles cannot have more than two j'oints common to both, without coinciding entirely. ■ ■"," If it be possihle, let ABC and ABE be two ©s which have more than two pts. in common, as A^ B, C. Join AB, BC: Then *.• -4 B is a chord of each circle, .*. the centre of each circle lies in the straight line, which bisects AB at right angles ; III. 1. and •.' BC is a chord of each circle, .*. the centre of each circle lies in the stuujrht line, wiiich bisects BC at right angles. III. 1. .'. the centre of each circle is the point, in which the two straight lines, which bisect AB and BC at right angles, meet. .•. the ©s ABC, ADE have a common centre, which is impossible ; III. 5 and 6. ,'. two © s cannot have more than two pts. common to both. Q. E. D. Note. We here insert two Propositions, Eucl. iii. 25 and IV. 5, which are closely connected with Theorems i. and x. of. this book. The learner should compare with this portion of the subject the note on Loci, p. 103. ' t .■ I" s \\l t34 EVCI.rtys ELEMENTS. [Book Itl. Proposition A. Problem. (Eucl. in. 25,) An arr of a circle hcii^j riven, to comylrte the circle of wl-ich it 18 a part. Let ^5C be the given arc. It is required to complete the © of which ABC is apa/rt Take B, any pt. in arc ABC, and join AB, BC From D :in.i E, the mitldie pts. of AB luul BC, draw DO, EO, ±s to AB, BC, uieeliny in 0. Then •.* AB is to be a chofd of the ©, .*. centre of the © lies in 1)0 ; III. 1. and *.' BC is to be a chord of the ®, .', centre of the © lies in EO. III. 1. Hence is the centre of the ® of which ABC is an arc, and if a © be described, with centre and radius OA, iliis will be the © requureil. Q. K. F. Book IIT.] PROPOSITION B. 135 Proposition B. Problem. (Eucl. iv. 5.) To describe a circle about a given triangle. m. Let ABC be the given A . It is required to describe a, © about the A. From D jind E, the middle pts. of AB and AC, draw DO, EO, ± s to AB, AC, and let them meet in 0. Then •/ AB is to be a chord of the 0, .•. centre of the © hes in DO. III. 1. And '.* AC is to be a chui-d of the ©, .•. centre of the © lies in EO. III. 1. Hence is the centre of the © which can be described aindit Ihe A , and if a © be described with centre and melius OA, this will be the © required. Q. £. F. Ex. If BAC be a ri lit angle, show that will coincide with the middle point of BC. :'>i ■^ " 4' i: ■ It. . .,, f 'I; 136 EUCLID'S ELEMENTS. [Book in. Proposition XI. Theorem. If one circle touch aether ivtcrnally at any point, the centre of the interior circle muat lie in that radius of the other circle which pasnes through that point of covtnct. Let the ADE touch the © ABC internally, and let A be a pt. of contact. Find the centre of © ABC, nnd join OA. The.i must the centre of © ADE lie in the radius OA. For if II t, let P be the centre of © ADE. Join OP, and produce it to meet the Oces in D and B. Then '.' P is the centre of © ADE, and from are drawn to the Oce of ADE the st. lines OA, OD, of which OD passes through P, .*. OD is greater than OA. III. 8, Cor. But O A = 0B; .'. OD is greater than OB, which is iniposiiible. /. the centre of © ADE is not out of the radius OA. .'. it lies in OA, Q. £• D, Book III.] PROPOSITION XII. li Proposition XII. Theorem. Ij two circles touch one another extervally at any jtoint, the straight line joining the centre of one toith that i oint of contact must ivhen produced pass through the centre of the other. Let © ABC touch © ADE externally at the pt. A. Let be the centre of © ABC. Join OA, and produce it to E. Then must the centre of © ADE lie in AE. For if not, let P be the centre of © ADE. Join OP uieeting the © s in J5, D ; and join AP. Then •.' OB=OA, and PD=AP,^ .'. OB nnd PD together=0^ and AP together ; .*. OP is not less than OA and AP together. But OP is less than OA and AP together, I. 20. which is i'iipossible ; .'. the centre of © ADE cmnot lie out of AE. Q. B. D. Ex. Three circles touch one another externally, whose centres are A, B, C. Shew that the difference between AB and AC is half as great as the difference between the diametert of tb© circles, whose centres are B and C. t"-j ii . II 138 EUCI.ins ELEMEl^TS. f.lool- TTi. Proi'osit.ov XIIT. Tiik()Ii::m. Ont circle cnnnol touch nnnfher at '.nore points than one, irhethcr it touch it internally or exterwyihj, First let the « ADE \u\\c\\ tlie Aii( inteiaallif iit pt. A. Then there can be no other poinl of contari. Take the centre of © ABC Then P, the centre ot © ADE, lies in OA. III. 11. Take any pt. E in the Qw of the © ADE, and //in OE. Then •.* from 0, a pt. within or without tlie ADS, two lines OA, OE are diavvn to the Q^'C, of which 0^ passes throuS ELKMENT^. [Book lit PROPosirroN XIV, Thkorem. E(pial chords in a cirrfe are equally distant from the centre ; and rntiversehjy thone which are equally dintant J'rof the centrt^ are equal to 07ie another. Let the chords AB, CD in the © ABDC be equal. Then must AB and CD be equally distant from the centre 0. Draw OP and OQ ± to AB and CD ; and join AO, CO. Then P and Q are the middle pts. of ^£ and Ci> : lU. 3. mdvAB=CD,.'.AP:=CQ. Then •/ AP=CQ, and AO^CO^ in the right-angled lb A OP, COQ, .'. OP^OQ ; I. E. Cor. p. 43. and .: AB and CD are equally distant from 0. Def. 8. Next, let AB and CD be equally distant from 0. Then must AB= CD. For ••• 0P= OQ, and A 0= CO, in the right-angled as AOP^ COQ^ .: AP=CQ, and .-. AB=CD. I. K. Cor. Q. B. D. Ex. In a circle, whose diameter is 10 inches, a chord is drawn, which is 8 inches long. If Hnotlier chord be drawn, at a distance of 3 inches from the centre, shew whether it is equal or not to the former. m. Book m.J PkOPOStTtOX XV. Hi Proposition XV. Thkorkm. The diameter is the greatest chord in a circk^ and of all others that which is nearer to the centre is alwaijs greater than one more remote ; and the greater in nearer to the centre than the ku. ♦ •. ■ > Let AB be a diameter of the ® ABDC, whose centre is 0, and let CD be any other chord, not a diameter, in the 0, nearer to the centre than the chord EF. TheA must AB he greater than CD, and CD greater than EF. Draw OP, OQ ± to CD and EF ; and join OC, OD, OE. Then •. • ^ 0= CO, and 0B= OD, I. Def. 13. .-. AB =mm of CO and OD, and .*. AB is greater than CD. I. 20. Again, •• • CD is nearer to the centre than EF, .'. OP is less than OQ. Def. 8. Now •.• sq. on 00= sq. on OE, ,*. sum ot'sqq. on OP, P0= sum of sqq on OQ, QE. I. 47. But sq. on OP is less than sq. on OQ ; .*. sq. on PC is greater than sq. on QE ; .•• PC is greater than QE ; and •. CD is greater than EF. Next, let CD be greater than EF. Then must CD he nearer to the centre than EF. For •.• CD is greater than EF, .'. PC is greater than QE. I i m 144 EUCLID'S ELEMENTS. [Book WT, I I Now the sum of sqq. on OV, PC=f'Uia of sqq. on 0^, QE. But sq. on PC is p-eater than sq. on (^E ; .'. sq. on OP is less than sq. on OQ ; .•, OP is less than 0^ ; and .'. CD is nearer to the centre than EF. Q. E. D. Ex. 1. Draw a chord of given length in a given circle, which shall be bisected by a given chord. Ex. 2. If two isosceles triangles be of equal altitude, and the sides of one be equal to the sides of the other, shew that their bases must be equal. Ex. 3. Any two chords of a circle, which cut a diameter in the same point and at equal angles, are equal to one another. Def. IX. A straight line is said to he a Tangent to, or to touch, a circle, lohen it meets and, behig prodticed, does not cut the circle. From this definition it follows that the tangent meets the circle in one point only, for if it met the circle in two points it would cut tlie (circle, since the line joining two points in the circumference is, being produced, a st'c;int. (III. 2.) Def. X. If from any point in a circle a line be drawn at right angles to the tangent at that point, the line is called a Normal to the circle at that point. Def. XI. A rectilinear figure is said to be described about a ;ircle, when each side of the figure touches the circle. And the circle is said to be inscribed in the figure. Book 11!. I PROPOSITION XVI. H3 PiioposiTiox XVI. Theorem. r^e straight line drawn at riyht avyles to the diaiiieter of a circle, from the extremity of it, is a tangent to the circle. lU ft ' III m JD P Let ABC be a ©, of which the centre is 0,and the diameter JOB Through B draw DE at right angles to AOB. I. 11. Tlien must DE he a tangent to the ® . Take any point F in DE, and join OP. Then, *.• z OBP is a liyht angle, .-. z OPJS is less than a right angle, and .-. OP is greater than OB. Hence P h a point without tlie ABC. Tn the sarin' way it may be shewn that every point in L or DE produced in either direction, except the point B, li 1.17. I. v.). Po without the DE is a tangent to the © . Def. &. Q. E. D. t44 EUCLID'S ELEMENTS. [Book lit Bo Proposition XVII. Problem. To draw a straight line from a given poi^it, either without or on tJ.ti circumference, which shall touch a given circle. Then and Let A be the given pt., without the BCD. Take the centre of © BCD, and join OA. Biseut OA in E, and with centre E and radius EO describe © ABOD, cutting the given ® in J5 and D. Join AB, AD. These are tangents to the © BCD. Join BO, BE. OE=-BE, .'. L OBE^ L BOE ; z J^jB=twice I OBE ; AE=BE, .'. L ABE= l BAE ; :.L OEB^tmceiABE ; .: sum ofzs AEB, OEB=twice sum ofzs Oli. that is, two right angles = twice z OB A ; .'. z OBA is a right angle, and .*. AB is a tangent to the © BCD. Similarly it may be shewn that AD is a tangent to © BCD. Next, Itt the given pt. be on the Qce of the ®, aa B. Then, if BA be drawn ± to the radius OB, BA is a tangent to the © at B. III. 16. Q. £. D. Ex. 1. Shew that the two tangents, drawn from a point with- out the circumference to a circle, are equal. Ex. 2. If a quadrilateral ABCD be described about a circle, shew that the bimu of AB nnd CD ia equal to the Punj of AD tOidBC. I. A. 1.32. I. A. I. .32. ABE, III. 16. ihi 111 Y nt fiook m.] PROPOSTTTON XVllI. I4S Proposition XV III. Theorem. If a straight line touch a circle, the straight line drawn from the centre to the point of contact must he 2 erpendicular to tJie line tov^ihing the cirele. Let the st. line DE touch the © ABC in the pt. 0. Find the centre, and join OC. Then must OC be ± to DE. For if it be not, draw OBFi. to DE, meeting the Oce in B. Then •." z OFC is a rt. angle, .*. /. OCF is less than a rt. angle, I. 17. and .'. OC is greater than OF. I. 19. But OC=OB, .'. OB is greater than OF, which is impossible ; .*. OF is not ± to DE, and in the same way it may he ihewn that no other line drawn from 0, but OC, is ± to DE , .-. 00 is ± to DE. Q. E. D. Ex. If two straight lines intersect, the centres of all circles "nT touched by both lines lie in two lines at right angles to each other. NoTK. Prop. XVIII. might be stated thus :—All radii of a circle are normals to the circle at the points where they meet the circwnferemc, ,, ^l!*-'^ ''t ! ■ 146 EUCLID'S ELEMENTS. [Book III. doot PUOPOSITION XIX. THEOnKM. If a straight line touch a circle, arid from the point of con- tact a straight line he drawn at right angles to the touching linCf the centre of the circle must he in that line. circ il Let the st. line DE touch the © ABC at the pt. C, and from C let CA be drawn 1 to DE. Then must the centre of the © he in CA. For if not, let F be the centre, and join FC. Then '.• DCE touches the ©, and FG is drawn from centre to pt. of contact, .-. z FCE is a rt. angle. III. 18. IJut I ACE is a rt. angle. .*. z FCE = L ACE, which id impossible. In the same way it may be shewn that no pt. out of CA can be the centre of tlie ® ; . .'. the centre of the ® lies in CA. Q. E. D. Ex. Two concentric circles being described, if a chord of the greater touch the less, the parts of the chord, intercepted between the two circles, are equal. Note, Prop. xix. miglit be stated thus :— Every normal to a circle pafises through the centre. III. dook nt] PnOPOSfTTON XX. HI Proposition XX. Theorem. The, angle at the centre of a circle in double of the angle at the circumference, subtended by the same arc. Let ABC be a © , the centre, BG any arc, A any pt. in the Oce. Then w ust iBOC= twice l BAG. First, sup]: «e to be in one of the lines containing the lBAC. Tlien •.• OA = 00, :.lOCA = lOAG; ,". sum of z s OGA, OAC = twice z GAG. Butz BOG = mm of z s OCA, OAC, • z BOO = twice z OAG. luax is, L BOG = twice z BAG I. A. 1.32. 11 h 14& pjfct.rr)\<; nip.MnKT^. [Book tit. Next, suppose to be within (fig 1), or without (fig. 2) the lBAG. Join AO^ and produce it to meet the Oce in D. Then, as in the first ca^je, L COD = twice z Gj.D, and z BOD = twice i BAD ; .*., fig. 1, sum of z s COD, BOD = twice sum of z s CAD, BAD, that is, z £00 = twice .' BAG. And, fig. 2, difference of z s COD, BOD = twice diflFerence of z 8 CAD, BAD, that is, z £0(7 = twice /. BAC. Q. E. D. Ex. From any point in a straight line, touching a circle, a straight line is drawn through the centre, and is terminated by the circumference ; the angle between tV-ese two straight lines is bisected by a straight line, which intersects the straight line joining their extremities. Shew that the angle between the lost two lines is half a light angle. m. the Book m.] NOTE IT. U 149 NoTB 2. On Flat and Reflex Angles. We have already explained (Note 3, T'ook I., p. 28) how Euclid's definition of an nn^fle niny be extended with advan- tage, so as to include the conception of an iinyle ecpiid to two right angles : and we now proceed to shew how the Definition given in that Note may be extended, so as to embrace angles greater than two right angles. I Ti 11 MD, rence Ircle, lated light light iveen Let WQ be a straight line, and QE its continuation. Then, by the Definition, the angle made by WQ and QE, which we propose to call a Flat Angle, is equal to two right angles. Now suppose QP to be a straight line, which revolves about the fixed point Q, and which at fiist coincides with QE. "When QP. revolving from right to left, c(tincides with QW, it has described an angle equal to two right angles. When QP has continued its revolution, fo as to come into the position indicated in the diagnun, it has described an angle EQP, indicated by the dotted line, greater than two right angles, and this we call a Reflex Angle. To assist the learner, we shall mark these angles with dotted lines in the diagrams. Admitting the existence of angles, equal to and greater than two right angles, the Proposition last proved may be extended, AS we pov/ procee4 tp ?hew, 1=: ill ISO EUCLID'S ELEMENTS. [Book in. Proposition 0. Theorem. Tha angle, not less than two right angles, at the cci, in of a circle is double of the angle at the circumference, subte)idid by the same arc. Fig. 1. Fig. 2. In the © ACBD, let the angles AOB (not less than two right angles) at the centre, and ADB at the cirr uniferer.ce, he subtended by the same arc ACB. TJien must l AOB = tvn(e l ADB. Join DO, and produce it to meet the arc ACB in C. Then :• ^ A 0C= twice z ADO, III. 20. and / 50(7= twice z BDO, III. 20. .*. sum of z s AOC, BOC= twice sum of z 8 ADO, BDO, that is, z ^05= twice z ADB. Q. E. D. Note. In fig. 1 , z ^ OB is drawn a flat angle, and in fig. 2, z AOB is drawn a reflex angle. Def. XII. The angle in a segment is the angh contained by two straight lines drawn from any point ip the arc tO the ex- U'euiiries of tb§ chord- ( fiook tit] I'ROrOSITICN XX'I. t.M n 1« 'II Proposition XXI. Tiikohkm. The angles in the same fcgment of a circle are equal to one another. "Let BAG, BDC !'<' iinuii's in the pame soL'nirnt BADC. Then. 7iivst I BAG= lB1)(\ First, when st'^iiient. BADC is fjieatei' tli:.-' i MMniciscle, From O, tlie centre, draw 0/>, OC. fFio;. 1.) Then, '. • z BOC= twice z /? J (\ I J I 20. i\m\ I BOC=t\\ue I BVl\ 111.20. ...iBAC= lBDC. Next, when seguicnt BADC is less than a s-en ici.clo, liCt E be the pt. of intcisectinn of JC, DB (Fitr. 2.) Then :■ i ABE= l DCE, by the Hist casr, mdiBEA^ I GED, I 1.5. ..iEAB=lET)C 1.32. that is, z £JC= z iJZ^C. Q. E. n. Ex. 1. Shew that, by assiiininf? the possi1ii!ity of an ai'i^Ie beinnr {jreater than two ri,<;ht angles, both the cases of thia ptoposition may be included in one. Ex. 2. If two sMaisrlit lines, whose extremities are in the circumference of a circle, cut one another, the triangles lornied by joining their extremities are equiangular to each other. '■■> .1 *: «<4 PiVcUb's k!MfE?^n. f Book m. Proposition XXII. Throrrm. The, opposite angles of any quadrilateral Jiyure, inncribed in a circle, are together eqrtal to two right angles. Let A BCD be a quadrilateral fig. inscribed in a ® . Then must ea>ch pair of its opposite l s he together equal to two rt. L 8. Draw the diagonals AC, BD. Then •.• z ADB= l ACB, in the same segment, III. 21. and z BDC= l BAC, in the sjime segment ; III, 21. /. sum of I s ADB, BDC=8mn o^ i s ACB, BAC \ that is, u ^DO=sum of z s ACB, BAC. Add to each z ABC. Then z 8 ADC, ABC toiTether=sum ofzs ACS, BA \ ABC; and .*. z s ADC, ABC together=two right i s. Similaily, it may be shewn, that I s BAD, BCD together=two right l ». I.; Q. E. D. Note. — Another method of proving this proposition is given on page 177. Ttt. Book III.j pkoposrttosr )t)(n. «53 Ex. 1. If one side of a quadrilateral figure inscribed in a circle be produced, the exterior angle is equal to the opposite angle of the quadrilateral. Ex. 2. If the sides AB^ DG of a quadrilateral inscribed in a circle be produced to meet in E, then the triangles EBGj EAD will be eqiiiaiigular. Ex. 3. Shew that a circle cannot be described about a rhombus. Ex. 4. The lines, bisecting any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet in the circumference of the circle. Ex. 6. AB^ a chord of a circle, is the base of an isosceles triangle, whose vertex is without the circle, and whose equal sides meet Che circle m D^ E \ shew that CD is equal to GE. Ex. 6. If in any quadrilateral the opposite angles be to- gether equal to two right angles, a circle may be described about that quadrilateral. Propositions xxiii. and xxiy., not being required in the method adopted for proving the subsequent Propositions in this book, are removed to the Appendix. Proposition xxv. has been already proved. Note 3. On the Method of Superposition, ae applied to Girclea. In Props, xxvi. xxvii. xxviii. xxix. we prove certain relations existing between chords, arcs, and angles in equal circles. As we shall employ the Method of Superposition, we must state the principles which render this method appli< cable, as a test of equality, in the case of figures with drculcr boundaries. '54 J.t CI.ID'S ELEMEXry. [rook lit. Dkf. XI 11. iJqual circlea arc those, of which the radii are tijual. For suppose ABC, A'B'C to he circles, of which tlie radii are equiil. Tlicii if A'Ji'i" h». applied to © ABC, ho tliiif O', tlie centre of A'B'C, coincides with O, tlio centre of Alii', it in evident that any jiartinilar point A' in the O^'e <>f the r.'rnier nmst coincide with .wiiic point A in Qro of tlio hitter, liecanse of the cqn.ility of tlie radii O'A' and OA. Hence Qw A'B'C must coincide with 0<'^' ATC, th,\'cis,& A'B'C =0 ABC. Further, when we have applied tlie circle A'I'(^' to the circle ABC, so that the centres coincide, Me may im ;;ine ABC to reniain fixed, while A'B'C revolves round the common centre. Hence \< e may suppose any particular point if' in the circumference of A'B'C to be maile to coincide with any par- ticular point B in the circumference of ^jBC. Apain, utiy radius O'A' of the circle A'B'C may be made to coincide with any radius OA of the circle ABC. Also, if A'B' and AB be equal arcs, they jnay be niade to coincide. Agiiin, every diameter of a circle divides the circle into equal segments. For let AOB be a diameter of the circle ^CBD, of which is the centre. Suppose the sejjment ACB to be ap- plied to the sejjrment ADB, so as to keep AB a common boundary : then the arc ACB must coincide with the arc ADB, because every point in each is equally distant from 0. Book III] rROPosmox xxvr. »55 Proposition XXVI. Tiikojirm. hi equal clrcli'A, thr (ti'dt, which unhtnul vipial autjlcif, vhither theij be at the centren or at the circumj'erences, vmst be equal. Let ABC, DEF be equal circles, niul l.'t i s LGO, EIIF ut their centres, and z s BAC, EDF at their O^'^^* ^^« equal. Then mud arc IiKC=arc ELF. lui, if © ABC he applied to DEF, . so that G coincides with II, and GB falls on HE, then, •.• GB=UE, .\ B will coincide with E. And ••• z BGC== z EIIF, .: GCMW fall on IIF ; and •.• GC=IIF, .'. G will coincide with F. Then •.' B coincides with E and C with F, .-. arc BKG will coincide with and be equal to aic ELF. Q. K. D. Cor. Sector BGCK is equal to sector EHFL. Note. This and the three following Propositions are, and will hereafter be assumed to be, true for the same eicle ?13 we]] {IS for equal circles, 156 EUCLID'S ELEMENTS [Book III. Proposition XXVII. Thkorem. In equal circles, the angles, which are subtended by equal arcs, whether they are at the centres or at the cvrcumferences, ruuJ be equal. Let ABC, DBF be equal circles, and let z s BGC, EHF ai their centres, and z s BAG, EDF at their Qc^s, be subtended by equal arcs BKC, ELF. Then must l BGC= l EHF, and iBAG=i EDF. For, if © ABC be applied to © DBF, so that G coincides with H, and GB falls on HE, then •.' GB^HE, .'. B will coincide with E ; and ••• arc BKC=&rc ELF, .'. C will coincide with F. Hence, GO will coincide with jHJ''. Then •/ £G cf»incide8 with E^^, and GO with HF, .'. L BGC will coincide with and be equal to z EHF. Again, •.• z BAC=ha\f of z BGC, III. 20. and z EDF=h»\{ of z EHF, III. 20. .-. z BAC= L EDF. I. Ax. 7. Q. E. D. Ex. 1. If, in a circle, AB, CD be two arcs of given magni- *Mde, and AC, BD be joined to meet in E, shew that the angle AEB is invariable. Ex. 2. The straight lines joining the extremities of the th »rils of two equal arcs of the same circle, towajrds the 3iune partSj are part|,Uel to eagb oth^Fr Book ni.] PROPOSITION XXVIII. 157 Proposition XXVIII. Theorem. In equal circles, the arcs, which are subtended hj equal chords, must be equal, the greater to the greater, and the less to the less. Let ABC, DEF be equal circles, and BG, EF eqnul chords, subtending the major arcs BAG, EDF, and the minor arcs BGG, EHF. Then must arc BAG= arc EDF, and arc BGG = arc EHF. Take the centres K, I, and join KB, KG, LE;LF. Then •.• KB=LE, mid KC=LF, and BC=EF, .-. z BKG = I ELF. I. a Hence, if ABG be applied to & DEF, so that K coincides with i, atid KB falls on LE, then •.• L BKG = /. ELF, .: KC will fall on LF ; and •.' KG = LF, .'. G will coincide with F. Then *.• B coincides with E, and G with F, .'. arc BAG will coincide with and be equal to arc EDF, and a,xc BGG EHF. Q. E. D. Ex. 1. If, in a circle ABCD, the chord AB be equal to the chord DG, AD must be parallel to BG. Ex. 2. If a straight line, drawn from A the middle point of an arc BG, touch the circle, shew that it is parallel to the chord BG. Ex. 3. If two equal chords, in a given circle, cut one an- other, the segments of the one shall be equal to the se«>meut9 *>' tJie other, each to e^ch» K8 r.UCLmS ELEMENTS. [Book III. Proposition XXIX. Theorem. 1% equal circles, the chords, which subtend equal arcs, must be equal. V M Let ABC, DEF be equal circles, and let BG, EF be chords subtending the equal arcs BGO, EHF. Then must chord BG = chord EF. Take the centres K, L. Then, if © ABG be applied to © DEF, so that K coincides with L, and B with E, and arc BGG falls on arc EHF, '.' arc BGG=aTc EHF, .'. G will coincide with F. Then '.* B coincides with E and G with F, ,% chord BG must coincide with and be equal to chord EF. Q. £. D. Ex. 1. The two straight lines in a circle, which join the extremities of two parallel chords, are equal to one another. Ex, 2. If three equal chords of a circle, cut one another in the same point;, wiihin the circle, that point is the centre. r. Book m.] MOTE 4. 1^0 6-t Note 4. On the Synwnetrical j)roperties of the Circle with regard to its diameter. The brief remarks on Symmetry in pp. 107, 108 may now he extended in the following way : A figure is said to be symmetrical with regard to a line, when every perpendicular to the line meets the figure at points which are equidistant from the line. Hence a Circle is Symmetrical with regard to its Diameter, because the diameter bisects every chord, to which it is per- pendicular. • i^ .i ds 1 ii the m Further, suppose AB to be a diameter of the circle iCBD, of which is the centre, and CD to be a chord perpendicular to AB. Then, if lines be drawn as in the diagram, we know that AB bisects (1.) The chord CD, III. I. (2.) The arcs GAD and CBD, III. 26. (3.) The angles CAD, COD, CBD, and the reflex angle DOC. I. 4. Also, chord C£== chord l)B, I. 4. and chord AC= chord A D. I. 4. 4 These Symmetrical relations should be carefully observed, because they are often suggestive of methods for the solution of problems. I> il iio kuctriy'^ EtEMkNn. fBooklft Peoposttton XXX. Problrm. To bisect a given are. Let ABC be the given arc. Tt is required to bisect the arc ABC. Join AC, and bisect the chord AC in D. From D draw DB± to AG. LIO. 1.11. Then will the arc ABC be bisected in B. Join BA, BG. Then, in ^s ADB, CDB, V AD^CD, and DB is common, and i ADB" i CDBy .-. BA = BC. I. 4. Rnt, in the same circle, the arcs, which are subtended by equal chords, are equal, the greater to the greater and the less to the less ; III. 28. and •.• DDy if produced, is a diameter, .*. each of the arcs BA, BC, is less than a semicircle, and .*. arc BA=avc BC. Thus the arc ABC is bisected in B. Q. R. F. Ex. If, from any point in the diameter of a semicircle, there be drawn two straight lines to the circumference, one to the bisection of the circumference, and the other at right angles to the diameter, the squares on these two lines are tot'cther double of the square on the radiut>. Book III.] PROnOSITTON XXXt. t6f Proposition XXXI. Thkorem. hi a circle, the angle in a semicircle is a right angle ; and the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle. i n 1, si ! '.I iiul Let ABC be a 0, its centre, and BC a diameter. Draw AC, dividing the © into the segments ABC, ADC. Join BA, AD, DC, AG. Then must the l in the seymcircle BAG be a rt. l , and l in segment ABC, greater than a semicircle, less than a rt. l , and l in segment ADC, less than a semicircle, greater than a rt. l . First, •.• BO=AO, .: iBAO= i ABO ; I. a. . •. z CO A = twice iBAO; I. 32. and •.• CO=AO, .'. l CA0.= l ACO ; I. a. .-. L BOA = twice l CAO ; I. 32. .'. sum of z s COA, 50J^ = twicesnm of z s BAG, CAO, that is, two right angles = twice z BAG. .'. L BAG is a right angle. Next, •.* z BAGis a rt. z , .'. z A BC is less than a rt. z . I. 17. Lastly, •.• sum of z s ABC, ADC=two rt. z a, III. 22. and z ABC is less^ than a rt. z , .'. z A DC is greater than a rt. z . (j. e. d, JJioTE. — For a simpler proof see page 178. 13 li^ ) id' ii V\ II t6i £:VCUD'S ELEMENTS. fBook lit Book III.] Ex. 1. If a circle be described on the radius of another circle as diiiraeter, any straight line, drawn from the point, where they meet, to the outer circumference, is bisected by the in- terior one Ex. 2. If a straight line be drawn to touch a circle, and be parallel to a chord, the point of contact will be the middle point of the arc cut ofiF by the chord. Ex. 3. If, from any point without a circle, lines be drawn touching it, the angle contained by the tangents is double of the angle contained by the line joining the points of contact, and the diameter drawn through one of them. Ex. 4. The vertical angle of any oblique-angled triangle mscribed in a circle is greater or less than a right angle, by the ingle contained by the base and the diameter drawn from the extremity of the base. Ex. 5. If, from the extremities of any diameter of a given circle, perpendiculars be drawn to any chord of the circle that Is not parallel to the diameter, the less perpendicular shall be equal to that segment of the greater, which is contained between the cii'cumference and the chord. Ex. 6. If two circles cut one another, and from either point jf intersection diameters be drawn, the extremities of these diameters and the other point of intersection lie in the same straight line. Ex. 7. Draw a straight line cutting two concentric circles, 20 that the part of it which is intercepted by the circumference of the greater may be twice the part int' rceptod by the circum- ference of the less. J/ a St a draiyh this line angles, n Draw tl Tat The Ag Ex. gents Book 1ft ler circle t, where the in- and be middle >e drawn ouble of contact, triangle e, by the from the a given ircle that shall be I between lier point of these the same c circles, mference e circum- Book III.] PROrOSTTION XXXIl. 163 Proposition XXXIL Tiieouem. Ij a atraiyht line touch a circle, and from ihc , oint of contact a atraiyht line he tlrawn. ciittiini the circle, the aiujla', made by this line with the line touch lu'j the circle must he equal to the angles, which are in the alternate seyments of the circle. Let the st. line AB touch the © CDEF in F. Draw the chord FD, dividiiig the® into segments FCD, FED. Then must l DFB= l in segment FCD, and L UFA = l in segment FED. From F draw the chord FC± to AB. Then FG is a diameter of the ® . III. 19. Take any pt. E in the arc FED, and join FE, ED, DC. Then •.' FDC ip a semicircle, .-. z FDC is a rt. z ; III. 31. .-. sum of z s FCD, CFD=a rt. z . I. 32. Also, sum of z s DFB, CFD=a rt. z . .-. sum of z s DFB, CFD =sum of z s FCD, CFD, and .-. z DFB== z FCD, that is, z DFB= z in segment FCD. Again, *.' CDEF is a quadrilateral iig. inscribed in a ®, . •. sum of z s FED, FCD = two rt. z s. III. 22. Also, sp.m of z s DFA, DFB^^two rt. z s. I. 13. .-. sum of z s DFA, DFB =s\\m of z s FED, FCD ; and z DFB has been proved = z FCD ; .'.lDFA^lFED, that is, z DFA— i in segment FED. Q. E. D. Ex. The chord joining the points pf contact of parallel taO' gents is a diameter, r^ i f i 'A \ 1 I : f . ! I M 164 EUCLID'S ELEMENTS. [Book in. Proposition XXXIII. Problem. On a given straight live to iJcMrihc a mjmeni of a circle capahle of containing an angle equal to a given angle. L Let AB he ihe given st. line, and C the given z . It is req^dred to describe on AB a segment of a © which shall contain an i — i G. At pt. A in St. line AB make a BAD= ^ C. I. 23. Draw AE± to AD, and bisect AB in F. From F draw FQ ±to AB, meeting AE in G. Join GB. 'J'lien 111 A s AGF, BGF ; •.• AF^BF, and FG is conmion, and z AFG= z BFG ; .-. GA = GB. I. 4. With G as centre and GA as radius describe a © ABH. Then will AHB be the-segment reqd. For *.' AD is± to !4£^,; a line passing through the centre, .-. AD is a tangent to the ABH. III. 16, And *.* the chord AB is drawn from the pt. of contact A, .: z BAD= L in segment AHB, III. 32. that is, the segment AHB contains an z = z 0, and it is described on AB, as was reqd. Q. E. F. Ex. 1. Two circles intersect in A, and throuj^h A is drawn a straight line meeting the circles again in P, Q. Prove that the angle between the tangents at F and Q is equal to the angle between the tangents at A. Ex. 2. From two given points on the same side of a straight line, given in position, draw two straight lines which shall conr tftiw a given angle, and be terminated in the given lipe, fiook in.] PROPOStTION XXXIV. 165 Proposition XXXIV. Problem. To cut off a, segment from a given circle, capable of con- taining an angle equal to a given angle. Let ABC be the given © , and D the given i . It is required to cut off from © ABC a segment capable of containing an l = l D. Draw the st. line EBF to touch the circle at B. At B make z FBC = iD. Then '.• the chord BC is drawn from the pt. of contact B, .'. L FBC = L in segment BAC, III. 32. that is, the segment BAC contains an z = lD\ and .'. a segment has been cut off from the © , as was reqrl. Q. E. F. Ex. 1. If two circles touch internally at a point, any straij^ht line passing through the point will divide the circles into f( -, ments, capable of containing equal angles. Ex. 2. Given a side of a triangle, its vertical angle, and the radius of the circumscribing circle : construct the triangle. Ex. 3. Given the base, vertical angle, and the perpendicular from the extremity of the base on the opposite side : construct the triangle. I i '\ I t % 1^6 kltCUD'S; KIEMEXTS. [Book ni. Proposition XXXV. Theorem. If two chords In a circle cut one another, the rectangle con- tained hy the segments of one of them, is equal to the rectangle contained by the segments of the other. Let the chords AC, BD in the A BCD Inteisect in the pt, P. Then must reel AP, PC=rfct. BP, PD. From 0, the centie, draw OM, ON ± s to AC, BD, smd join OA, OB, OP. Then *.* AC is divided equally in Af and unequally in P, .*. rect. AP, PC with sq. on AfP=sq. on AM. II. 5. Adding to each the sq. on MO, rect. AP, PC with sqq. on MP, MO=sqq. on AM, MO ; .'. rect. AP, PC with sq. on OP =sq, on OA. I. 47. In the same way it may be shewn that rect. BP, PD with sq. on OP=sq. on OB. Then *.• sq. on OA = sq. on OB, :. rect. AP, PC with sq. on 0P= rect. BP, PD with sq. ou OP ; .-. rect. AP. PC=rect. BP, PD. q. b. d. Ex. 1. A and B are fixed points, and two circles are described passing throuj^h them ; PCQ, P CQ' are chords of these circles intersecting in C, a point in AB ; shew that the rectangle CP, CQ is equal to the rectangle CP', CQ'. Ex. 2. If through any point in the common chord of two circles, which intersect one another, theie be drawn any two other chords, one in each circle, their four extremities shall all lie in the circumference of a circle. Book m.] PROPOSITION XXXVI. 167 Proposition XXXVI. Theorem. 7/, from any 'point without a circle, two straight linen he draioii, one of which cnt: the circh', and the other touclut it; the rectamjJe cantainrd hy the whole Jitie wliih r»'ii tie circlCf and the part of it without the circhy tnud 1 e equa^ to the square on the line which touches it. Let D be any pt. without the © ABC, and let the st. lines DBA, DC be drawn to cut and touch the © , Then must rect. AD, DB=sq. on DC. From 0, the centre, draw OM bisecting AB in M, and join OB, OC, OD. Then *.• AB is bisected in M and produced to D, /. rect. AD, DB with sq. on MB=sq. on MD. II. 6. Adding to each the sq. on MO, rect. AD, DB with sqq. on MB, ilfO=sqq. on MD, MO. Now the angles at ilf nnd G are rt. z s ; III. 3 and 18. .'. rect. AD, DB with sq. on 05= sq. on OD ; .-. rect. AD, DB with sq. on 05= sqq. on 00, DC. 1. 47. And sq. on 0£=sq. on 00 ; '. rect. AD, PJ5=sq. on DC, Q. e, d. , ltd a lf3 EUCLID'S ELEMENTS. [Book ni. Proposition XXXVII. Theorem. If ^ from a point without a circle, tliere he drawn two straight linen, one of which cuts the circle, and the other meets it ; if the rectanyle contained by the whole line which cuts the circle, and the //art of it without the circle, be equal to the square on the line wnich meets it, the line which meets muot touch the circUk Let ^ bo a pt. without the © BCD, of which is the centre. From A let two st. lines A(W, AB be dniwn, of which ACD cuts the © and AB meets it. Then if red. DA, AC=sq. on AB, AB mtist touch the ®. Draw AE touching the © in E, and join OB, OA, OE. Then '.• ACD cuts the ®, and AE touches it, .-. rect. DA, J.O=sq. on AE. III. 36. But rect. DA, AC=sq. on AB ; Hyp. .*. sq. on ^B=sq. on AE ; .-. AB=AE. Then in the as OAB, OAE, \' OB—OE, and OA is common, and AB=AE, .:lABO=- lAEO. But ^ ^^0 is art. z ; .•. z ABO is a rt. z . Now BO, if produced, is a diameter of the ® ; .'. AB touches the 0. I. c. III. 18. III. 16. Q. E. P. ni. Book HLJ MISCELLANEOUS EXERCISES. 169 Miscellaneous Exercises on Book III, 1. The segments, into which a circle is cut by any Btnii^'ht line, contain angles, whose difference is equal to the incliniitioii to each other of the straij^ht lines touching the circle at the ex- tremities of the straij^ht line wiiich divides the circle. 2. If from the point in which a nutuber of circles touch each other, a straight line be drawn cuttmg all the circles, shew that the lines which join the points of intersection in each circle with its centre will be all parallel. 3. From a point Q in a circle, QN is drawn perpendicular to a chord PP', and QM perpendicular to the tangent at P : shew that the triangles NQF, QPM are equiangular. 4. ABf AC are chords of a circle, and D, E are the middle points of their arcs. If DE be joined, shew that it will cut off equal pans from AB, AG, 5. One angle of a qiiadrilateral figure inscribed in a circle is a right angle, and from the centre of the circle perpendiculars are drawn to the sides, shew that the sum of their squares is equal to twice the square of the radius. 6. A is the extremity of the diameter of a circle, any point in the diameter. The chord which is bisected at sub- tends a greater or less angle at A than any other chord through 0, according as and A are on the same or opposite sides of the centre. 7. If a straight line in a circle not passing through the centre be bisected by another and this by a third and so on, prove that the points of bisection continually approach the centra of the circle. 8. If a circle be described passing through the opposite angles of a parallelogram, and cutting the four sides, and the points of intersection be joined so as to form a hexagon, the straight lines thus drawn shall be parallel to each other. 9. If two circles touch each other externally and any third oiroLB touch both, prove that the difference of the distaaoes of 170 EUCLTjyS ELEMENTS. [Book III. the centre of the third circle from the centres of the other two is invariable. 10. Draw two concentric circles, such that those chords of the outer circle, which touch the inner, may equal its diameter. 11. If the sides of a quadrilateral inscribed in a circle be bisected and the middle points of adjacent sides joined, the circles described about the triangles thus formed are all equal and all touch the original circle. 12. Draw a tangent to a circle which shall be parallel to a given finite straight line. 13. Describe a circle, wliich shall have a given radius, and its centre in a given straight line, and shall also touch anothei straight line, inclined at a given angle to the former. 14. Find a point in the diameter produced of a given circle, from which, if a tangent be drawn to the circle, it shall be equal to a given straight line. 16. Two equal circles intersect in the points A, B, and through B a straight line CBM is drawn cutting them again in G, M. Shew that if with centre C and radius BM a circle be described, it will cut the circle ABC in a point L such that arc ilL=arc AB. Shew also that LB is the tangent at B. 16. AB is any chord and AG Si, tangent to a circle at A ; GDE a line cutting the circle in D and E and parallel to AB. Shew that the triangle AGD is equiangular to the triangle EAB. 17. Two equal circles cut one another in the points A. H ; BC is a chord equal to AB ; shew that AG is a tangent to the other circle. 18. A, B are two points ; with centre B describe a circle, such that its tangent from A shall be equal to a given line. 19. The perpendiculars drawn from the angular points of a triangle to the opposite sides pass through the same point Book 111.] MtSCELLANBOVS EXERCISES. l?l 20. If perpendiculars be dropped from the angular points of a triangle on the opposite sides, shew that the sum of the squares on the sides of the triangle is equal to twice the sum of the rectangles, contained by the perpendiculars and that part of each intercepted between the angles of the triangles and the point of intersection of the perpendiculars. 21. When two circles intersect, their common chord bisects their common tangent. 22. Two circles intersect in A and B. Two points C and D are taken on one of the circles ; 0-4, CB meet the other circle in E, F, and DA, DB meet it in (?, H : shew that FG is parallel to EH. 23. A and B are fixed points, and two circles are described passing through them ; CP, CP' are drawn from a point G on AB produced, to touch the circles in P, P' ; shew that CP=CP'. 24. From each angular point of a triangle a perpendicular is let fall u^on the opposite side ; prove that the rectangles con- tained by the segments, into which each perpendicular is divided by the point of intersection of the three, are equal to each other. 25. If from a point without a circle two equal straight lines be drawn to the circumference and produced, shew that they will be at the same distance from the centre. 26. Let 0, 0' be the centres of two circles which cut each other in A, A'. Let B, B' be two points, taken one on each circumference. Let C, C be the centres of the circles BA B\ BA'B'. Then prove that the angle CBC is the supplement of the angle OA'(y. 27. The common chord of two circles is produced to any point P ; PA touches one of the circles in A ; PBG is any chord of the other : shew that the circle which passes through A, B, G touches the circle to which PA is a tangent. 28. Given the base of a triangle, the vertical angle, and the length of the line drawn from the vertex to the middle point of the base : construct the triangle. a >•• iji ^VCLWi, ELEMENTS. [fiook XSL 29. If a circle be described about the triangle ABC^ and a straight line be drawn bisecting the angle £^(7 and cutting the circle in D, shew that the angle BOB will be equal to half the angle BA C. 30. If the line AD bisect the angle A in the triangle ABC, and BD be drawn without the triangle making an angle with £0 equal to half the angle BAC, shew that a circle may be described about ABCD. 31. Two equal circles intersect in A,B: PQT perpendicular to AB meets it in Tand the circles in P, Q. AP, BQ meet in B ; AQ, BP in S ; prove that the angle BTS is bisected by TI. . '■' - ' 32. If the angle, contained by any side of a quadrilateral and the adjacent side produced, be equal to the opposite angle of the quadrilateral, prove that any side of the quadrilateral will subtend equal angles at the opposite angles of the quadrilateral. 33. If DE be drawn parallel to the base BC oi a, triangle ABC, prove that the circles described about the triangles ABC and ADE have a common tangent at A. 34. Describe a square equal to the difference of two given squares. 35. If tangents be drawn to a circle from any point without it, and a third line be drawn between the point and the centre of the circle, touching the circle, the perimeter of the triangle formed by the three tangents will be the same for all positions of the third point of contact. 36. If on *:he sides of any triangle as chords, circles be de- scribed, of which the segments external to the triangle contain angles respectively equal to the angles of a given triangle, those circles will intersect in a point. 37. Prove that if ABC be a triangle inscribed in a circle, such that BA = BG, and A A' be drawn parallel to BC, meeting the circle again in A\ and A'B be joined cutting ACin E, BA touches the circle described about the triangle AEA'. 38. Describe a circle, cutting the sides of a given square, so that its circntnference may be divided at the points of inteiv section into eight equal arcs. Book m.] MlSCMLtA^om £)CMPCISES. m a Iff If 39. AB is the diameter of a semicircle, D and E any two points on its circumference. Shew that if the chords joining A and B with L> and E, either way, intersect in F and G, the tangents at D and E meet in the middle point of the line FG, and that FG produced is at right angles to AB. 40. Shew that the square on the tangent drawn from any point in the outer of two concentric circles to the inner equals the difference of the squares '^n the tangents, drawn from any point, without both circles, to the circles. 41. If from a point without a circle, two tangents PT, PT\ at right angles to one another, be drawn to touch the circle, and if from T any chord T(^ be drawn, and from T' a perpen- dicular T'M be dropped on TQ, then T'M^QM. 42. Find the loci : (1.) Of the centres of circles passing through two given points. (2.) Of the middle points of a system of parallel chords in a circle. (3.) Of points such that the difference of the distances of each from two given straight lines is equal to a given straight line. (4.) Of the centres of circles touching a given line in a given point. (5.) Of the middle points of chords in a circle that pass through a given point. (6.) Of the centres of circles of given radius which touch a given circle. (7.; Of the middle points of chords of equal length in a circle. (8.) Of the middle points of the straight lines drawn from a given point to meet the circumference of a given circle. r 43, If the base and vertical angle of a triangle be given, find the locus of the vertex. 44. A straight line remains parallel to itself while one of its extremities describes a ciicle. What is the locus of the other extremity ? f?4 JStTCL/D'S ELEMENTS. tfiooknt 45. A ladder slips down between a vertical wall and a horizontal plane : what is the locus of its middle point? 46. ABC is a line drawn from a point A, without a circle, to meet the circumference in B and 0. Tangents are drawn to the circle at B and G which meet in D. What is the locus of£>? 47. The angular points A,C of a parallelogram ABCD move on two fixed straight lines OA, OC, whcse inclination is equal to the angle BCD ; sheAv that one of the points B, D, which is the more remote from 0, will move on a fixed straight line passing through 0. 48. On the line AB is described the segment of a circle in the circumference of which any point C is taken. If A C, BG be joined, and a point P taken in AG so that GP is equal to CB, find the locus of P. 49. The centre of the circle CBED is on the circumference of ABD. If from any point A the lines J 1>0 and -^ii"!) be drawn to cut the circles, the chord BE is parallel to CD, 50. If a parallelogram be described having the diameter of a given circle for one of its sides, and the intersection of it? diagonitis on the circumference, shew that the extremity of each of the diagonals moves on the circumference of anothei ^circle of double the diameter of the first. 51. One diagonal of a quadrilateral inscribed in a circle is fixed, and the other of constant length. Shew that the sides will meet, if produced, on the circumferences of two fixed circles. Book in.i Eucuns proof of in. 23. m We here insert Euclid's proofs of Props. 23, 24 of Book III. first observing that he gives the following definition of similar segments : — Dbf. Similar tsegmenls of circles are those in whith the angles are equal, or which contain equal angles. Proposition XXIII. Theorem. Upon the same straight line, and upon the .« '"me side of it, there cannot be two similar segments of circles, not coinciding with each other. 'Ul *-H If it be possible, on the same base AB, and on the same side ■>f it, let there be two similar segments of ©s, ABC, ABD, which do not coincide. Because © ADB cuts © ACB in pts. A and B, they cannot cut one another in any other pt., and .'. one of the segments must fall within tlie other. ^(to\ Let ADB M\ within ACB. Draw the st. line BDC and join CA , DA. Then •.* segment ADB is similar to segment ACB, .-. I ADB= L ACB. Or the extr. z of a A =the intr. and opposite z , which ia ) (\i) impossible ; " ,', the segments gtinnot but coincide. (^ fi. o. i 170 EUCLID'S ELEMENTS. I «• • [AOUiL ill. Proposition XXIV. Thkorfm. Similar segments of circles, upon equal straight lines, are zqual to one another. Let ABC, DEF be similar segments of © s on equal st. lines AB,DE. Then must segment ABC = segment DEF. For if segment ABO be applied to segment DEF, so that A may be oti D nnd AB on DE, then B will coincide with E, and AB with DE ; .'. segment ABG must also coincide with segment DEF ; ' III. 2S. .♦. segment J.£C= segment DEF. Ax. 8. Q. E. D. We gave one Proposition, C, page 150, as an example of the way in which the conceptions oi Flat and Reflex Angles may be employed to extend and simplify Euclid's proofs. We here give the proofs, based on the sirae conceptions, of t}ie impoi- tant propositions xxii. and xx:u. Book III.] ANOTHER PROOF OF III. 22. 177 Proposition XXII. Theorem. 'Hie. opposite angks of any quadrilateral figure, inscribed in a circle, are together equal to two right angles. Let A BCD be a quadrilateral fig. inscribed in a ®. Then nmst each pair of its opposite l s he together equal to irwo rt. is. From 0, the centre, draw OB, OD. Then •.' i BOD =twice l BAD, III. 20. and the reflex z DOU= twice z BCD, III. C. p. 150. •, sum of z s at 0= twice sum of z s BAD, BCD Eu5 sum of z s at 0=4 right z s ; I. 15, Cor. 2. .-. twice sum of z s BAD, BCD =4 right z s ; .-. sum of z s BAD, BCD=two right z s. Similarly, it may be shewn that sum of z s ABC, ADC=two right z p. Q. £. Db S)i » 178 EUCLID'S ELEMENTS. [Book III. Proposition XXXI. Theorem. In a circle, the angle in a semicircle is a right angle ; and the angle in a segment gveatir than a semicircle in /cxs- than a right angle ,' and the angle in a segment less than a semicircle ia ■jreati'- than a right angle. Let ABC be a ©, of which is the centre and BO a «liaineter. • Draw AC, dividing the © into the segments ABC, ADC. Join BA, AD, DC. Then must the l in the semicircle BAC be a rt. i , and i in segment ABC, gi eater than a semicircle, less than a rt. i , and l in segment ADC, le-is than a semicircle, greater than a rt. 1 .. First, •.• the flat angle £00= twice z BAC, IH. C p. 150. .'. / BACis a rt. / . Next, •.• z BAC is a rt. z , '. z ABC is less than a rt. ^ . I. 17. Lastly, *.• sura of z s ABC, ADC =t\vo rt. z s, III. 22. and z ABC is less than a rt. z , ,*. I ADC is greater than a rt. z . Q. E. D. »■- '^v-^,^ ,«„,tto.t^c >."-"'•'-- • 1.- Se.;^ O't' ')CvCi ft ~ ' *" ' « ■ .^-^i •-Jan* Vr C<_^ (v d ^oi'xV u - afc) -Vc ■-"^^'". ■