H^^ IN MEMORIAM FLORIAN CAJORl c^ CD. In either case, the tract BD or DB, between the two ends of the tracts, whose beginnings coin- cide, is called the difference of the two tracts, and we are said to subtract the one from the other. Ordinarily we mention the greater tract first in speaking of difference. 19. The symbols of addition and subtraction are -\- and — (plus and minus), thus : AB+ CD=AD and AB - CD = BD. It is important to note here the order of the letters. In summing a number of tracts, as AB, CD, EF, etc., to KL, BCD K Fig. 8. we have AB + CD ^- EF-- -\- KL = AL (Fig. 8) . The order of the summands is indifferent, and this important fact is called the Commutative Law of Addition. Thus AB^ CD\EF=AB^EF-\- CD=EF-\-AB-\- CD, etc. 18 GEOMETRY. 20. When beginning and end of a tract or of any mag- nitude are exchanged, the tract or magnitude is said to be reversed, and the reverse is denoted by the sign — . Thus the reverse of AB is BA, or AB = — BA. If we add a magnitude and its reverse, the sum is o, or AB + {-AB) = AB + BA = o. The same result o is obtained by subtracting, from a magni- tude, itself or an equal magnitude; and, in general, it is plain that to subtract CD yields the same result as to add (Fig. 9) the reverse DC. The reverse of a magnitude is A B CD c A B Fig. 9. often called its negative, the magnitude itself being called its positive. Similar rules hold for adding and subtracting arcs of a circle or of equal circles. ANGLES. 21. The indefinite extent of a ray on one side of a point O, as OA, is called a half-ray : it has a beginning O, but no end. Two half-rays, OA and 0A\ which together make up a whole ray, are called opposite or counter (Fig. 10). Now let two half-rays, OA and OB, have the same be- ginning O ; the opening or spread between them is a mag- nitude : it may be greater or less. Suppose OA and OB to be two very fine needles pivoted at O ; then OB may fall exactly on OA, or it may be turned round from OA ; and INTRODUCTION. 19 the amount of turning from OA to OB^ or the spread between the half-rays, is called the angle between them. We may denote it by a Greek letter, as a, written in it ; or by a large Roman letter, as (9, at its vertex (where the half- rays meet) ; or by three such letters, as A OB, the middle Fig. io. one being at the vertex, the other two anywhere on the half- rays. The symbol for angle is ^. 22. The angle is perfectly definite in size, it has two ends or boundaries ; namely, the two half-rays, sometimes called arms. When we would distinguish these arms as beginning and end, we mention the letter on the beginning- arm first, and the letter on the end-arm last ; thus, AOB ; here OA is the beginning and OB the end of the angle. Exchanging beginning and end reverses the angle ; thus, BOA = -AOB. 23. Two angles whose ends or arms may be made to fit on each other simultaneously are named equal ; they are also congruent. Two angles whose arms will not fit on each other simultaneously are unequal; and that is the less angle whose end-arm falls within the other angle when their beginnings 20 GEOMETRY, coincide; the other is the greater; thus, AOB> AOC (Fig. II). 24. We sum angles precisely as we sum tracts ; we lay off a, p, etc., around (9, making the end of each the begin- ning of the next : the angle from first beginning to last end Fig. II. is the sum. So, too, in order to subtract ^ from a, lay off /8 from the beginning towards the end of a ; the angle from the end of /? to the end of a is the difference, a — (i. Or we may add to a the reverse (or negative) of /? : the sum will be « + (— y8) or a — ^ (Fig. 12).^ 1 It is important to note the close correspondence of tract and angle: the former is related to points as the latter is to rays (or half-rays). The tract is the simplest magnitude that lies between points, that distinguishes them and keeps them apart; likewise the angle is the simplest magnitude that lies between rays (in a plane), that distinguishes them and keeps them apart. So, too, we define equality and inequality among tracts and among angles, quite similarly, and without being compelled before- hand to form the notion of the size either of a tract or of an angle. We may now define the distance between two points to be the tract between them, and the dz's- tance between two (half-) rays to be the angle between them, leaving for future decision which tract and which angle if there should prove to be several. INTRODUCTION. D 21 Fig. 12, AXIOMS. 25. At this stage we must recognize and use certain dic- tates or irresoluble facts of experience, called axioms. {'k^ni)imv[\G^2in?> something worthy, like the Latin dignitas ; in fact, older writers use dignity in the sense of axiom. But Euclid's phrase is Kotvat Iwomi — common notions.^ Some have no special reference to Geometry, but pervade all of our thinking about magnitudes ; such are (i) Things equal to the same thing are equal to each other. (2) If equals be added to, subtracted from, multiplied by, or divided by, equals, the results will be equal. 22 GEOMETRY. (3) If equals be added to or subtracted from unequals, the latter will remain unequal as before. (4) The whole equals, or is the sum of, all its distinct parts, and is greater than any of its parts. (5) If a necessary consequence of any supposition is false, the supposition itself is false. Others concern Geometry especially, as : (6) All planes are congruent. (7) Two rays can meet in only one point. The extremely important axiom (7) may be stated in other equivalent ways, thus : Two rays cannot meet in two or more points ; or. Two rays cannot have two or more points in common ; or. Only one ray can go through two fixed points ; or, A ray is fixed by two points. 26. A statement or declaration in words is called -3, prop- osition. The propositions with which we have to deal state geometric facts and are also called Theorems {Oeoiprjfxa, from OewpcLv, to look at, means the product of mental contempla- tion). Propositions are often incorrect; theorems, never. Subordinate facts, special cases of general facts, and facts immediately evidenced from some preceding facts, are called Corollaries or Porisms {Tvopiafxa = deduction). We may now proceed to investigate lines and angles, and find out what we can about them. The first and simplest things we can learn concern Th. 1.] CONGRUENCE, ?3 CONGRUENCE. 27. Theorem I. — All rays are congruent. Proof. Let L and V be any two rays (Fig. 13). On L take any two points, A and B ', on L' take any two points, A' and B', so that the tract AB shall equal the tract A'B'. Think of Z and L' as extremely fine rigid spider-threads, and in thought place the ends of the tract AB on the ends of the tract A'B', A on A', and B on B'. B A B Fig. 13. Then A and A' become one and the same point, and B and B' become one and the same point ; through these two points only one ray can pass (by Axiom 7) : hence Z and Z', which go through these two points, now become one and the same ray ; that is, they fit precisely, they are con- gruent. Quod erat demonstrandum = which was to be proved = oTTcp iSa Set^at, — the solemn Greek formula; whereas the Hindu, appealing directly to intuition, merely said Paqya — Behold ! 28. In the foregoing proof we assumed that on any ray we could lay off a tract equal to a given tract, or that on any ray we could find two points, A and B^ as far apart as two other points. A' and B\ This assumption that something can be done, is called a Postulate (atrry/xa), i.e. a demand, which must be granted before we can proceed further. Actually to carry out the construction, we need a pair of compasses. 24 GEOMETRY. [Th. II. 29. Theorem II. — If two points of a ray lie in a certain plane, all points of the ray lie in that plarie. Proof. Regard the surface of paper or of the blackboard as a plane, and suppose it covered with a fine rigid film, itself a plane. Let L be any ray having two points, A and B, in this plane. Through these two points suppose a second plane drawn or passed; by definition (Art. 13) it will intersect our first plane, or film, along a ray /; this ray / goes through the two points, A and B, and lies wholly (with all its points) in the first plane ; also the ray L goes through A and B, and only one ray can go through the same two points, A and B, by Axiom 7 ; hence L and / are the same ray ; but / has all its points in the first plane ; hence L has all its points in the first plane, q. e. d. Query : What postulate is assumed in this proof ? Corollary. If a ray turn about a fixed point P^ and glide along a fixed ray Z, it will trace out a plane (Fig. 14). Fig. 14. For it will always have two points — namely, the fixed point and a point on the fixed ray — in the plane drawn through the fixed point and the fixed ray. Th. II.] CONGRUENCE. 25 Query : What postulate is here implied ? — Henceforth it is understood that all our points, lines, etc., are complanar, i.e. lie in one and the same plane. 30. In the foregoing Theorem and Corollary we observe clauses introduced by the word if. Such a clause is called an Hypothesis, i.e. a supposition. The result reached by reasoning from the hypothesis and stated immediately after the hypothesis, is called the Conclusion. 31. All logical processes consist in one or both of two things : the formation of concepts, as of lines, surfaces, angles, etc., and the combination of these concepts into propositions. Geometric concepts are remarkable for their perfect clearness and precision — we know exactly what we mean by them ; this cannot be said of many other concepts, about which diverse opinions prevail, as in Political Economy. Hence it is that Geometry offers an unequalled gymnasium for the reason or logical faculty. We shall now generate some new concepts. Let the student note their definiteness as well as the mode of their formation. 32. Let OA and OB be any two co-initial half- rays, forming the angle A OB. Think of OA as held fast and of OB as turning about the pivot O, starting from the position OA. As it turns (counter-clockwise), the (Fig. 15) angle A' AOB increases. Finally, let it return to its original posi- tion, OA ; then the whole amount of turning from the upper 26 GEOMETRY. [Th. III. side of OA back to the under side of OA, or the full spread around the point O, is called a ///// angle (or round angle, or io? = computation = thinking together). The first two propositions are called premisses, the third and last, in which the other two are thought together, is called conclusion. All reasoning may be syllogized, but this is rarely done, as being too formal and tedious. 38. Theorem VI. — Two counter half- rays bound a straight angle. A O Fig. 20. For, let OA and OA^ be two such counter half-rays (Fig. 20) forming the whole ray AA\ Turn the upper half of the plane film round O as pivot until the upper OA^ falls on the lower 0A\ then, since the ray is reversible, the ray AA^ will fit exactly on the ray A^A ; i.e. the two angles AOA' and A'OA are congruent and equal; and the two compose the round angle A OA ; hence each is half of AOA ; i.e. each is a straight angle, q. e. d. 39. Theorem VII. — Conversely, The half-rays bounding a straight angle a7'e counter. P Fig. 21. Let OA and OA^ bound a straight angle (Fig. 21) AOA' ; also let TB and TB' be two counter half-rays ; then they 30 GEOMETRY. [Th. VII. bound a straight angle BPB\ by Theorem VL Since all straight angles are congruent, we may fit these two on each other; i.e. we may fit OA and OA^ on PB and PB^ ; but BB^ is a ray ; so then is AA^ ; i.e. OA and OA^ are counter, q. e. d. 40. We may define a straight angle as an angle bounded by counter half-rays. Then we may prove Theorem V. thus : The ends of all straight angles are pairs of counter half- rays (or form whole rays) ; But all such pairs (or whole rays) are congruent (by Theorem I.) ; Therefore, all ends of straight angles are simultaneously congruent. But when the ends of angles are (simultaneously) congru- ent, so are the angles themselves. Hence all straight angles are congruent, q. e. d. Here the first conclusion, introduced by " therefore," is deduced from two premisses ; but the second, introduced by "hence," is apparently deduced from only one. Only apparendy, however ; for one premiss was understood but not expressed ; namely, all straight angles are angles ivhose ends are congruent. Without some such implied additional premiss, it would be impossible to draw the conclusion. Such a maimed syllogism, with only one expressed premiss, is called an enthymeme. The great body of our reasoning is enthymematic. We shall frequently call for the suppressed premiss or reason by a parenthetic question (Why?). 41. Now draw two rays, LV and MM\ meeting at O. Each divides the round angle about O into two equal straight angles, and together they (Fig. 22) form four angles a, /3, a', (^'. Two angles, as a and /?, that have a common arm, are called adjacent. Accordingly we see at once : Th. IX.] CONGRUENCE. 31 Theorem VIII. — Where tuio fays intersect, the sum of two adjacent angles is a straight angle. Fig. 22. Two angles whose sum is a straight angle are called supplemental ; two angles whose sum is a round angle we may call explemental. Two angles as a and «', the arms of the one being counter to the arms of the other, are called opposite, or vertical, or counter. Theorem IX. — Whoi two rays meet, the opposite angles formed are equal. For a-{- (3 = S (a straight angle) (why?) ; and «'H-y8 = S (why?). Hence a -{- /3 = a' -{- /3 (why?) ; therefore a = a'. Simi- larly let the student show that jS = /3'. Q. e. d. An important special case is when the adjacents, a and /?, are equal. Each then is half of a straight angle, and there- fore one fourth of a round angle ; and each is called a right angle. Now let the student show that if a — /?, then «' = y8 and a = /3', or Corollary. When two intersecting rays make two equal adjacent angles, they make all four of the angles equal (Fig. 23)- Def. Rays that make right angles with one another are called normal (or perpendicular) to each other. N.B. The normal relation is mutual. How ? 32 GEOMETRY. [Th. IX. Def. Two angles whose sum is a right angle are called complemental. iy Fig. 23. 42. Are we sure that through any point on a ray we can draw a normal to the ray? Let O be any point on the ray LV (Fig 24). Let any half-ray, pivoted at O, start Fig. 24. from the position OL and turn counter-clockwise into the position 0L\ At first the angle on the right is less than the angle on the left, at last it is greater ; the plane, the turning, and the angle are all continuous ; hence in passing from the stage of being less to the stage of being greater, it passes Th. X.] CONGRUENCE. ZZ through the stage of equaUty. Let OR be its position in this stage ; then ^LOR = ^ ROL' ; i.e. OR is normal to LV. Moreover, in no other position, as OS^ is the ray normal to LL' ; for LOS is not = LOR unless (9»S falls on ORf but is less than L OR when OS falls within the angle LOR, while SOL' is greater than LOR ; hence LOS and SOL' are not equal ; i.e. OS is not normal to LL' when OS falls not on OR. Similarly, when L OS is greater than L OR. Hence Theorem X. — Through a point on a ray one, and only one, ray ca7i be drawn normal to the ray. 43. Def. A ray through the vertex of an angle, and forming equal angles with the arms of the angle, is called the inner Bisector or mid-ray of the angle. The inner bisector of an adjacent supplemental angle is called the outer bisector of the angle itself. Thus OL bisects innerly and OE bisects outer ly the angle AOB (Fig. 25). Exercise. Prove that there is one and only one such inner mid-ray. 34 GEOMETRY. [Th. XL 44. Theorem XI. — The inner Bisector of an angle bisects also its explement innerly. Proof. Let 6>/ bisect 1^ AOB innerly; then '^AOI= ^ lOB; call each a; then a+BOI'=a + AOr (why?) ; take away a ; then BOV=AOr (why?) ; i.e. the ray //' bisects innerly the angle BOA, the explement of A OB. Show that the angles marked a' are equal. 45. Theorem XII. — The inner and outer Bisectors of an angle are normal to each other. Proof. Let 01 and OE bisect (Fig. 25) innerly and outerly the angle A OB. Then, by definition, the angles marked a are equal, and the angles marked ^ are equal ; also the sum of -{- a -\- a -\- ^ -\- fi = S ; hence a-{- [i=^S; or, lOE = a right angle, q. e. d. TRIANGLES. 46. Thus far we have treated only of rays intersecting in a single point. But, in general, three rays Z, M, N (Fig. Fig. 26, 26) will meet in three points, since each pair will meet in one point, and there are three pairs : {MN) , {NL) , {LM) . Th. XIIL] TRIANGLES. 35 Denote these points by A, B, C. Then the figure formed by these three rays is called a triangle, trigon, or three-side. A, B, C are its vertices ; a, /3, y its inner angles ; BC, CA, AB, its inner sides, or simply its sides. Its angles and sides are called its parts. It is the simplest closed rectihnear figure, and most important. If instead of taking three rays we take three points A, B, C, then we may join them in pairs by rays; and since there are three pairs, BC, CA, AB, then there are three rays, which we may name Z, M, N. Thus we see that three points determine three rays, just as three rays determine three points. This equivalent deter- mination of the figure by the same number of points as of rays makes the figure unique and especially important. We denote it by the symbol A. We now ask, When are two triangles congruent? 47. Theorem XIII. — Tim A having two sides and the included angle of the one equal respectively to two sides and the included angle of the other are congruent. The data are: Two A, ABC and A^B^C, having the three equalities, AB = A^B\ -AC^A'C, « = cc' (Fig. 27). Fig. 27. Proof. Fit the angle n on the angle a' ; this is possible, because the angles are equal and congruent. Then A falls 36 GEOMETRY. [Th. XIV. on ^' ; also the point B falls on B^ (why? Because AB = A'B'), and C falls on C (why?). Hence the three ver- tices of the two A coincide in pairs ; therefore the three sides of the two A coincide in pairs (why ? Because through two points, as A {A') and B {B'), only one ray can pass). Q. E. D. Corollary i. The other parts of the two A are equal or congruent in pairs of correspondents : ^ = ;8', y = y', BC=B'C\ Corollary 2. Pairs of equal parts lie opposite to pairs of equal parts. 48. Theorem XIV. — Two A having two angles and the included side of the one equal respectively to two angles and the included side of the other are co figment (Fig. 28). Fig. 28. Data : Two AB=:A'B'. A ABC, A'B'C, having « = «', (3 = (S\ Proof. Fit AB on A' B' ; this is possible (why?). Then a will fit on a' (why?), and y8 on /?' (why?) ; i.e. the ray ^C will fit on A'C, and the ray BC on B'C. Then the point C will fall on C (why? Because two rays meet in only one point) ; i.e. the two A fit exactly, q. e. d. Th. xvi.j TRIANGLES, 37 49. We may now use the conditiofis of congrueitce thus far estabUshed to generate new notions that may be used in estabHshing other Theorems. Def. The ray normal to a tract at its mid-point is called the mid-normal of the tract. Theorem XV. — Any point on the mid-normal of a tract is equidistant from its ends (Fig. 29). PaC Data : AB a tract, M its mid-point, L the mid-normal, F any point on it. Proof. Compare the A AFAf and BFM. We have AM^BM (why?). FM^FM, ^ AMF='4. BMF (why?) ; hence the A are congruent (why?) ; and FA = FB. Q. E. D. Def. A A with two equal sides, like AFB, is called isosceles ; the third side is called the base, and its opposite angle the vertical angle. 50. Theorem XVI. — The angles at the base of an isosceles A are equal; and conversely. 38 GEOMETRY. [Th. XVI. Data: ABC an isosceles A, AB its base, AC and BC its equal sides (Fig. 30). Fig. 30. Proof. Take up the A AB C, turn it over, and replace it in the position BCA. Then the two A ACB and BCA have the equal vertical angles, C and C, also the side AC ~ BC (why?) and BC = AC (why?) ; hence they are con- gruent (why ?) , and the '^A = '^B. q. e. d. Conversely, A A whose basal angles are equal is isosceles. Let the student conduct a proof quite similar to the fore- going. Def. The ray through a vertex and the mid-point of the opposite side is called the medial of that side. Corollary i . In an isosceles A the medial of the base is normal to it, and is the mid-ray of the vertical angle. Corollary 2. When the medial of a side of a A is normal to the side, the A is isosceles. Prove it. Corollary 3. When the medial of a side bisects the opposite angle, the A is isosceles. Can you prove it ? Th. XVI.] LOGICAL DIGRESSION. 39 LOGICAL DIGRESSION. 51. When the subject and predicate of a proposition are merely exchanged, the proposition is said to be converted, and the new proposition is called the cotiverse. Thus X is Y; conversely y Y\% X. In general, converses of true prop- ositions are not true, but false. Thus, The horse is an animal is always correct, but The animal is a horse is generally false. A proposition remains true after simple conversion only when subject and predicate are properly quantified, thus: All horses are some animals; conversely. Some ani7nals are all horses. Both propositions are correct and mean the same thing. But they are awkward in ex- pression, and such forms are rarely or never used. When the quantifying word is all or its equivalent, the term is said to be taken universally ; when it is some or its equiva- lent, the term is said to be taken particularly. Thus in the foregoing example horse is taken universally, but animal particularly. The only useful conversions are of proposi- tions in which both subject and predicate are universal. In the great body of propositions only the subject is quantified universally, the quantifier is omitted from the predicate, but a particular one is understood. To show that a universal quantifier is admissible requires in general a distinct proof. 52. In order to convert an hypothetic proposition, we exchange hypothesis and conclusion. Thus, '\iX is F, Uvs, V; the converse is, if 6^ is V, X is K All such hypothetic propositions may be stated categorically ^ thus : All cases of X being Y are cases of U being V ; conversely. All cases of U being V are cases of X being Y. This converse is plainly false except when the quantifier all is admissible in the first predicate. 40 GEOMETRY. [Th. XVII. 53. But while the converse of a true hypothetic propo- sition is generally false, the contrapositive is always true. This latter is formed by exchanging hypothesis and conclu- sion and denying both. Thus : If Jf is F, then U \% V \ contrapositive, If U is not V, then X is not K Or, if a point is on the mid-normal of a tract, then it is equidistant from the ends of the tract ; contrapositive, If a point is not equidistant from the ends of a tract, then it is not on the mid-normal of the tract. 54. Theorem XVII. — An outer angle of a t^ is greater than either inner non-adjacent angle. Data: Let ABC be any A, a' an outer angle, /3' a non- adjacent inner one (Fig. 31). Proof. Draw the medial CM and lay off MD = MC ; also draw AD. Then in the A AMD and BMC we have AM=BM {why}), MD = MC (why?), and ^AMD = ^BMC (why?) ; hence the A are congruent (why?), and ^MBC^-^ MAD (why ?) . But ^ MAD is only part of the ^ «' ; hence «'> ^ MAD (why?) ; i.e. a'>(3'. Q. e. d. Similarly, prove that a' > y. Th. xviil] triangles. 41 55. Theorem XVIII . — If t7vo sides of a A are utiequal, then the opposite angles are unequal in the same sense {i.e. the greater angle opposite the greater side) (Fig. 32). Fig. 32. Data: ABC a A, AC>AB, AR the mid- ray of the angle at A, AB' laid off = AB. Proof. ABR and AB^R are congruent (why?) ; hence ^ ABR = ^ AB^R (why ?) ; but ^ AB^R > C (why ?) ; i.e. '^.ABC>^ACB. Q. e. d. Conversely, If two angles of a A are unequal, the opposite sides are unequal in the same sense. Proof. The opposite sides are not equal ; for when the sides are equal, the opposite angles are equal (Theorem XVI.), and contrapositively, when the angles are unequal, the opposite sides are unequal. Then, by the preceding Theorem, the greater angle Hes opposite the greater side. 56. Join BB^ ; then AR is the mid-normal of BB' (why ?), and hence angle CBB^=z angle BB'R (why ?) . Hence angle BB'C>B'BC (why?); hence BC>B'C (why?). But BC=AC-AB; htnct BC> AC - AB ; i.e. 42 GEOMETRY. [Th. XIX. Theorem XIX. — Any side of a A is greater than the difference of the other two. Add AB to both sides of this inequahty and there results AB + BC>AC\ i.e. Theorem XX. — Any side of a A is less than the sum of the other two. This fundamental Theorem is here proved on the sup- position that AB ^ C, it would need no formal proof. 57. Theorem XXI. — A point not on the mid-normal of a tract is not equidistant from the ends of the tract. Data : AB the tract, MN the mid-normal, Q any point not o^MN {Y\g. 33). Fig. 33. Proof. Draw QA and QB ; one of them, as QA, must cut MN^X. some point, as/^. Then QB< QP^PB (why?), and PB = PA (why ?) ; hence QB < QP+ PA ; i.e. QB < QA. Q. E. D. Of what Theorem is this the converse ? If now we seek for a point equidistant from A and B, we can find it on the mid-normal of AB and only there ; hence the locus of a point equidistant from the ends of a tract is the mid-normal of the tract. Th. XXIII.] TRIANGLES. 43 58. Theorem XXII. — Two A with the three sides of the one equal respectively to the th7'ee sides of the other are congruent. Data: ABC and AB'C the two A, and AB = A'B\ BC=^B^C\ C^= C'^' (Fig. 34). Fig. 34. Proof. Turn the A A'B'C over and fit AB^ on AB so that O shall fall (say) at D. Draw CD. Then A and B are on the mid-normal of CD (why?) ; hence the ray AB is the mid-normal of CD (why?) ; hence the angle CAB =1 angle DAB, and angle CBA = angle DBA (why ?) . Hence the A are congruent (why?), q.e.b. N.B. As to when the A must be turned round and when turned over, see Art. 94. 59. Theorem XXIII. — A. J^rom any point outside of a ray one normal may be drawn to the ray. Data: Pthe point, LV the ray (Fig. 35). Proof. From P draw a ray far to the left, as PA, making the angle PAL > angle PAD. Now let the ray turn about /* as a pivot into some position far to the right, making angle PAL < PAD. The plane, the angle, the motion, all being continuous, in passing from the stage of being unequal GEOMETRY. [Th. XXIII. in one sense to the stage of being unequal in the opposite sense, the angles made by the moving ray with the fixed ray must have passed through the stage of equality. Let PN be Fig. 35. the ray in this position so that angle PNL = angle PNV ; then each is a right angle by Definition, and /Wis normal to LL\ Q. E. D. B. There is only one ray through a fixed point and normal to a fixed ray. Proof. Any other ray than PA^, as PD, is not normal to LV ; for the outer angle PDL is > the right angle PND (why ?) . Q. E. D. C. The normal tract PN is shorter than any other tract from P to the ray LL' . Proof. For the right angle at iV^is > angle PDN (why ?) ; hence PN < PD (why ?) . q. e. d. D. E. Equal tracts from point to ray meet the ray at equal distances from the foot of the normal; and conversely. Proof. For, if DPD^ be isosceles, then the normal /Wis the medial of the base (why?). F. Two, and only two, tracts of given length can be drawn from a point to a ray. Th. XXIV.] TRIANGLES. 45 Proof. For two, and only two, points are on the ray at a given distance from the foot of the normal. G. Of tracts drawn to points unequally distant from the foot of the normal^ the one drawn to the remotest is the longest. Proof. In the A PDA, angle PDA > PAD (why?) ; hence PA > PD (why ?) . q. e. d. Similarly, PA' > PD. H. Equal tracts from the point to the ray make equal angles with the normal from the point to the ray and also equal angles with the ray itself ; and conversely. I. Of unequal tracts from the point to the ray, the longest makes the greatest angle with the normal and the least with the ray. Let the student conduct the proof of H and /. 60. Theorem XXIV. — Two A having two angles and an opposite side of one equal respectively to two angles and an opposite side of the other are congruent. Data : ABC and A'B'C two A having AB = A'B\ angle a = angle a', angle y = angle y' (Fig. 36). Fig. 36. 46 GEOMETRY, [Th. XXIV. Proof. Fit a' on « ; then B' falls on B (why ?) , and A^ C falls along A C. Draw the normal BN. Then B C and B C make the same angle, y = y', with the ray AN; hence they are = and meet the ray in the same point (why ?) ; i.e. C falls on C ; i.e. the A are congruent. Q. e. d. 6i. We now come to the so-called ambiguous case, of two A with two sides and an opposite angle in one equal to the two sides and the corresponding opposite angle in the Fig. 37. other. Let ABC and A' B' C (Fig. 37) be the two A, with AB = A'B\ BC = B'C, and angle a = angle a'. Fit a' on Th. XXV.] TRIANGLES. 47 a ; then AB^ falls on AB, B' on B ; but since from a point B {B') we may draw two equal tracts to the ray AL, the side B' C may be either of these equals and may or may not fall on BC. In general, then, we cannot prove congruence in this case. But if ^C be > AB, then angle « > angle y (why?), and there is only one tract on the right of AB drawn from B to the ray AC and equal to ^C; the other tract equal to BC must be drawn outside of AB and to the left. Hence in this case, when the angle lies opposite the greater side, the A are congruent. Hence Theorem XXV. — Two A having two sides and an angle opposite the greater in one equal to two sides and an angle opposite the greater side in the other are congruent. Corollary. Two right A having a side and any other part of one equal to a side and the corresponding part of the other are congruent. Fig. 38. 62. We have seen (Art. 47) that when two A have two sides and included angle in one equal to two sides and 48 GEOMETRY. [Th. XXVI. included angle in the other, they are congruent. But what if the included angles are not equal ? Let ABC and A^B^ C be the two A, having AB^AB\ BC= B'C, but ji > (i'. Slip the upper film of the plane along until A^B' fits on AB and let C fall on D. Draw the mid-ray BM of the angle CBD, let it cut AC ^.i M, and draw DM. Then the A CBMd^vA DBMdiTQ congruent (why?) ; hence AM-\- MB = AC (why?), and AC>AI?, or AOA'C. Hence Theorem XXVI. — Two A having two sides of one equal to two sides of the other, but the included angles unequal, have also the third sides unequal, the greater side lying opposite the greater angle. Conversely, Two A having two sides in one equal to two sides in the other, but the third sides unequal, have the included angles also unequal, the greater angle being opposite the greater side. Proof. The included angles are not equal; for if they were equal, the A would be congruent (why?) and the three sides would be equal. Hence the included angles are unequal, and the relation just established holds ; namely, the greater angle lies opposite a greater side. q. e. d. 63. Theorem XXVII. — Every point on a mid-ray of an angle is equidistant from its sides. Data : O the angle, MM' the mid-ray, /*any point on it. Proof. From P draw the normals PC and PD ; they are (Fig. 39) the distances of P from the ends of the angle. Then the APOC and POD are congruent (why?) ; hence PD = PC. Q. E. D. Th. XXVIIL] TRIANGLES. 49 Conversely, A point equidistant from the ends of an angle is on a mid-ray of the angle (Fig. 39). Fig. 39. Proof. If FC=^FD, then the A FOC and FOB are congruent (why ?) ; hence angle FOD = angle FO C. Q. e. d. Accordingly we say that the mid-rays of an angle are the locus of a point equidistant fro fn its ends. *64. It is just at this stage in the development of the doctrine of the Triangle that we are compelled to halt and introduce a new concept before we can proceed any further. The necessity of this step will appear from what follows (which may, however, be omitted on first reading, at the option of teacher or student) . Def Two A not congruent are called equivalent when they may be cut up into parts that are congruent in pairs. Theorem XXVIII. — Any A is equivalent to another A having the sum of two of its angles equal to the smallest angle of the given A. Data : ABC the A, a the least angle (Fig. 40). 50 GEOMETRY, [Th. XXIX. Proof. Through M, the mid-point oi BC, draw AM a.nd make MD = MA. Then the A A CM and DBM are con- gruent (why?), the part AMB is common to ABC and ABD, and the sum of the angles ADB and ^^Z) = angle BAC. Q.E.D. B Fig. 40. Corollary. The sum of the angles in the new A is equal to the sum of the angles in the old A. * 65. We may now repeat this process, applying it to the smallest angle, as A, of the A ABD. In the new A ABE the smallest angle, as A, cannot be greater than \ of the original angle a in ABC ; after n repetitions of this process we obtain a A, as ALB, in which the sum of the angles A and L cannot be > — of the original angle a in the A 2" ABC. By making n as large as we please, we make — as small as we please, and so we make — of angle a smaller than any assigned magnitude no matter how small. Meantime the other angle B has indeed grown larger and larger, but has remained < a straight angle. Hence the sum of the angles in the A ALB cannot exceed a straight angle by any amount however small; but the sum of the angles in ALB = sum of the angles in ABC ; hence Theorem XXIX. — 77ie sum of the angles in any A can- not exceed a straight angle by any finite amount. Th. XXX.] TRIANGLES. 51 Corollary i. The outer angle of a A is not less than the sum of the inner non-adjacent angles. Corollary 2. From any point outside of a ray there may be drawn a ray making with the given ray an angle small at will. Proof. From P draw any ray PA, and lay off AB = PA (Fig. 41). Then the angle PBA is not greater than Fig. 41. \PAN (why?); now lay off BC = PB (why?); then angle PCB is not > ^ angle PBA (why?) ; proceeding this way, we obtain after n constructions an angle PLN not > — of the angle PAN, and by making n large enough we may make this — as small as we please, q. e. d. 2** *66. Theorem XXX. — If the sum of the angles in any A equals a straight angle, then it equals a straight angle in every A (Fig. 42). 52 GEOMETRY. [Th. XXX. Hypothesis: ABC a A A^-B-\-C^S. with the sum of its angles Proof, (i) Draw any ray through C, as CD. Then if the sum of the angles in the A ACD and BCD \i^ S—x and S—y^ x and y being any definite magnitudes however small, then on adding these sums we get 2S—{x-^y); and on subtracting the sum, S, of the supplemental angles at Z> we get S — {x-\-y) for the sum of the angles of the A ABC. Now if this sum be S, then x and y must each be O ; i.e. the sum of the angles in each of the A A CD and BCD is S. Now draw DE and DF\ in each of the four small A the sum of the angles is still =6". (2) We may now make a A as large as we please and of any shape what- ever, but the sum of the angles will remain = S. For, take the same A ABC, and draw CD normal to AB. Then the sum of the (Fig. 43) angles in the A ACD is ^S", as has Fig. 43. been shown above ; also angle Z> is a right angle ; hence the angles A and ACD are complementary. Now along A C fit another A A CD^ congruent with A CD ; then all the angles of the quadrilateral ADCD^ are right, and the figure is called a rectangle. Now we can place horizontally side by side as many of these rectangles, all congruent, as we please, say / of them ; we can also place as many of them vertically, one upon another, as we please, say q of them ; Th. XXX.] TRIANGLES. 53 and we can then fill up the whole figure into a new rectan- gle, as large as we please. About each inner junction-point of the sides of the rectangles there will be four right angles plainly. Now connect the two opposite vertices, as A and Z, of this rectangle. So we get two congruent right A, in each of which the sum of the angles is S. Then any A that we cut off from this right A will, by the foregoing, have the sum of its angles equal to S. Since p and q are entirely in our power, we may make in this way any desired right A and from it cut off any desired oblique A, with the sum of its angles = S. q. e. d. Hence either no A has the sum of its angles = S, or every A has the sum of its angles = S. 67. A logical choice between these alternatives is impos- sible, but the matter may be cleared up by the following considerations : Across any ray ZM draw a transversal T, cutting LM at O, and making the angles a, jS, y, 8. Through any point, as O'j of 7" draw a ray (Fig. 44) L'M' making angle «' = a. Fig. 44. - This is evidently possible (why ?) . Then plainly /8' = ^, y' = y, 8' = 8, a' = a ; they are called corresponding angles ; 5+ GEOMETRY. [Th. XXX. also a and y', /? and 8' are equal, — they are called alternate angles ; also a and 8', as well as /8 and y', are supplemental, — they are called interadjacent angles. 68. Now let /*be the mid-point of 00^ ; on it as a pivot turn the whole right side of the plane round through a straight angle until O falls on (9', and (9' falls on O. Then, since the angles about O and O^ are equal as above stated, the half-ray OL will fall and fit on the half-ray O^M\ and the half-ray O^V on the halt-ray OM. Accordingly, if the rays LM and VM^ meet on one side of the transversal T, they also meet on the other side of T. 69. Three possibilities here lie open : (i) The rays ZJ/and Z'J/' may meet on the left and also on the right of T, in different points. (2) They may meet on the left and ajso on the right of T, in the same point. (3) They may not meet at all. No logical choice among these three is possible. But in all regions accessible to our experience the rays neither converge nor show any tendency to converge. Hence we assume as an Axiom A. Two rays that make with any third ray a pair of corresponding angles equal, or a pair of alternate angles equal, or a pair of interadjacent angles supplemental, are non-inter sectors. 70. But another query now arises. Is it possible to draw another ray through 6>' so close to V that it will not meet OL however far both may be produced? Here again it is impossible to answer from pure logic. An appeal to experi- ence is all that is left us. This latter testifies that no ray Th. XXX.] PARALLELS. 55 can be drawn through (9' so close to (9'Z' as not to approach and finally meet the ray OL. Hence we assume as another Axiom B. Through any point in a plane only one non- intersector can be drawn for a given straight line. This single non-intersector is commonly called the parallel, through the point, to the straight line. 71. It cannot be too firmly insisted, nor too distinctly understood, that the existence of any non-intersector at all, and the existence of only one for any given point and given ray, are both assumptions, which cannot be proved to be facts. The best that can be said of them, and that is quite good enough, is that they and all their logical consequences accord completely and perfectly with all our experience as far as our experience has hitherto gone. Even then, if there be any error in our assumptions, we have thus far been utterly unable to find it out. A geometry that should reject either or both of these assumptions would have just as much logical right to be as the geometry that accepts them, and such geometries lack neither interest nor importance. They may be called Hyper- Euclidean in contradistinction from this of ours, which from this point on is Euclidean (so-called from the Greek master, Euclides, who distinctly enunciated the equivalent of our Axioms in a Definition and a Postulate). Note. — Observe the relation of Axioms A and B : the one is the converse of the other. Observe also that the necessity of assuming the first lies in our igno- rance of the indefinitely great, and the occasion of assuming the other lies in our ignorance of the indefinitely small. See Note, Art. 301. 72. Accepting our Axioms as at least exacter than any experiment we can inake, we may now easily settle the ques- 56 GEOMETRY. [Th. XXXI. tion as to the sum of the angles in a A. Let ABC be any A ; through the vertex C draw the one parallel to the base AB. Then a=a\ (3 = ft' (why?) ; also a' + y -^ ft' = S ; hence a-\-y + ft = S; i.e. (Fig. 45) Theorem XXXI. — The sum of the angles in a A. is a straight angle. Fig. 45. Corollary i. The outer angle E equals the sum of the inner non-adjacent angles a and y (why?). Corollary 2. If two angles of a A be known, the third is also known. Corollary 3. If two A have two angles, or the sum of two angles of the one equal to two angles, or the sum of two angles of the other, then the third angles are equal. Corollary 4. To know the three angles of a A is not to know the A completely, for many A may have the same three angles. Such A are similar, as we shall see, but are not congruent ; they are alike in shape, but not in size. 73. Next to normahty, parallehsm is the most important relation in which rays can stand to each other, and we must now use the new relation in the generation of new concepts. Th. XXXIII.] PARALLELOGRAMS. 57 Theorem XXXII. — Parallel Intercepts between parallels are equal. Data : L and L\ J/ and M , two pairs of parallels (Fig. 46). Fig. 46. Proof. Draw BD. Then the A ABD and CDB are congruent (why?), and AB = CD, BC= DA. q. e. d. Def. The figure ABCD formed by two pairs of parallel sides is called a parallelogram, and may be denoted by the symbol O. A join of opposite vertices, as BD, is called a diagonal. 74. Theorem XXXIII. — Properties of the parallelogram. A. The opposite sides of a parallelogram are equal. This has just been proved. B. The opposite angles of a parallelogram are equal. Proof. « = y8 (why?); p=a' (why?); hence « = «'. Q. E. D. Corollary. Adjacent angles of a parallelogram are sup- plementary. C. Each diagonal of a parallelogram cuts it into two con- gruent A. Prove it. 58 GEOMETRY. [Th. XXXIV. D. The diagonals of a parallelogram bisect each other (Fig. 47)' FIG. 47. Proof. The A AMB and CMD are congruent (why?) ; hence AM^ CM, BM= DM. q. e. d. 75. We may now convert all the foregoing propositions and obtain as many criteria of the parallelogram. Theorem XXXIV. — A'. A 4-side with its opposite sides equal is a parallelogram. Data : AB = CD, AD=^ CB (Fig 48). Fig. 48. Proof. Draw BD. Then ABD and CDB are congruent (why?); hence ;8 = 8 ; and AD and CB are parallel; similarly, AB and CD are parallel; hence ABCD is a par- allelogram. Q. E. D. Th. XXXIV.] PARALLELOGRAMS. 59 B'. A 4-side with opposite angles equal is a parallelogram. Data: (^ = a', y8 + /S' = y + y' (Fig. 49)- ^^-^i Fig. 49. Proof. Since a = «', /? + y = ^' + y' (why ?) . Hence ^ = y'j ;S' = y ; z.. the three medials pass through O. q. e. d. Corollary. Each medial cuts off a third from each of the other two. For C0 = 20R (why?). Def. The point of concurrence of the medials is called the centroid of the A. It is two-thirds the length of each medial from the corresponding vertex. 86. Theorem XLIV. — The mid-normals of the sides of a A concur. Data : ABC a A, Z and J/ mid-normals to the sides BC and CA, meeting at S. Th. XLV.] CONCURRENTS. 67 Proof. 6* is equidistant from B and C (why?), and from Cand A (why?) ; hence S is equidistant from A and B (why?), or is on the mid-normal of AB (why?) ; hence the mid-normals concur (Fig. 58). Q. e. d. Fig. 58. Corollary. S is equidistant from A, B, and C, and no other point in the plane is (why?). Def. The point of concurrence of the mid-normals is called the circumcentre of the A. 87. Def. A tract from a vertex of a A normal to the opposite side is called an altitude of the A. Sometimes, when length is not considered, the whole ray is called the altitude. Theorem XLV. — The altitudes of a A concur. Proof. Using the preceding figure, draw the A A^B^C. Its sides are parallel to the sides of ABC (why?) ; hence its altitudes are the mid-normals Z, J/, N; and these have just been found to concur. Also, since ABC may be any A, A'B'C may be any A ; hence the altitudes of any A concur, q. e. d. 68 GEOMETRY. [Th. XLVI. Def. The point of concurrence of altitudes is called the orthocentre (or alticentre) of the A. Def. In a right A the side opposite the right angle is called the hypotenuse ( = subtense = under-stretch) . Queries : Where do circumcentre and orthocentre lie : (i) in an acute-angled A ? (2) in an obtuse-angled A ? (3) in a right A ? 88. Theorem XLVI. — The inner mid-rays of the angles of a A concur. Data : ABC a A, AL, BM, CN the inner mid-rays of its angles (Fig. 59). Fig. 59. Proof. Let AL and BM intersect at /. Then / is equi- distant from AB and AC, and from AB and BC (why?) ; hence / is equidistant from AC and BC \ hence / is on the inner mid-ray of the angle C ; i.e. the three inner mid- rays concur in /. q. e. d. Def The point of concurrence of the inner mid-rays is called the in-centre of the A. Th. XLVIL] exercises L 69 89. Theorem XLVII. — The outer mid-rays of two angles and the inner mid-ray of the other angle of a A concur. Let the student conduct the proof (Fig. 60). Fig. 60^ Def. The points of concurrence are called ex-centres of the A : there are three. EXERCISES I. Little by little the student has been left to rely more and more upon his own resources of knowledge and ratiocination in the conduct of the foregoing investigations. He has now possessed himself of a large fund of concepts, and he must test his ability to wield, combine, and manipulate them in forging original proofs of theorems. Let him bear always in mind the fundamental logical principle that every example 70 GEOMETRY. of a general concept has all the marks of thai general concept. Let him begin his proof by stating precisely the data, the given or known facts, let him draw a corresponding diagram in order to have a clearer view of the spatial relations in- volved, let him note carefully what concepts are present in the proposition, let him draw auxiliary lines and introduce auxiliary concepts at pleasure. But let him exhaust simple means before trying more complicated, let him distinguish, by manner of drawing, the principal from the auxiliary rays, and especially let him be systematic and consistent in the literation of his figures. 1. How many degrees in a straight angle? In a right angle ? Historical Note. — For purposes of computation the round angle is divided into 360 equal parts called degrees, each degree into 60 equal minutes (partes minutcB primse), each minute into 60 equal seconds (partes minutae secundce), denoted by °, ', " respectively. This sexagesimal division is cumbrous and unscientific, but is apparently permanently established. It seems to have originated with the Baby- lonians, who fixed approximately the length of the year at 360 days, in which time the sun completed his circuit of the heavens. A degree, then, as is indicated by the name, which means s:tep in Latin, Greek, Hebrew {gradus, ^ad/xos (or T/Mrjfxa), ma'a/ak),wa.s primarily the daily s^ep of the sun eastward among the stars. The Chinese, on the other hand, determined the year much more exactly at 365^ days, and accordingly, in defiance of all arithmetic sense, divided the circle into 365^ degrees. 2. The angles of a A are equal ; how many degrees in each? Remark. — Such a A is called equiangular, more commonly equi- lateral, but better still regular. 3. Show that this regular A is equilateral. 4. One angle of a A is a right-angle ; the others are equal ; how many degrees in each ? EXERCISES I. 71 5. One angle of a A is twice and the other thrice the third ; what are the angles ? 6. Two angles of a A are measured and found to be 46° 37' 24" and 52° 48' 39"; what is the third? 7. One angle of a A is measured to be 61° 22' 40"; the others are computed to be 49° 34' 28" and 69° 2' 43" ; what do you infer? 8. A half-ray turns through two round angles counter- clockwise, then through half a right-angle clockwise, then through a straight angle counter-clockwise, then through \ of a round angle counter-clockwise, then through ^ of a straight angle clockwise ; what angle does it make in its final position with its original position ? 9. 6^ is a fixed point (called origin) on a ray, A and B are any pair of points, M their mid-point. Show and state in words that 2 0M= OA + OB. 10. A, B, C are three points on a ray, A\ B\ C are mid-points of the tracts BC, CA, AB, and O is any point on the ray ; show that 0A-{- 0B-\- OC=OA^+ OB^ + OC. 11. A, B, C, D, O are points on a ray; A\ B\ C are mid-points of AB, BC, CD) A", B", are mid-points of A'B', B'C ; J/ is the mid-point of A"B" ; prove SOM= OA + sOB-\-sOC+OZ>. 12. What are the conditions of congruence in isosceles A ? In right A ? 13. In what A does one angle equal the sum of the other two? Def. A number of tracts joining consecutively any number of points (first with second, second with third, etc.) is called a broken line, or train of tracts, or polygon. Where the last 72 GEOMETRY. point falls on the first the polygon is closed; otherwise it is open. Unless otherwise stated, the polygon is supposed to be closed. The points are the vertices, the tracts are the sides of the polygon. The closed polygon has the same number of vertices and sides, and we may call it an n-angle or n-side. The angles between the pairs of consecutive sides are the angles of the polygon, either inner or outer ; unless otherwise stated, inner angles are referred to. Inner and outer angles at any vertex are supplemental. When each inner angle is less than a straight angle, the polygon is called convex ; otherwise, re-entrant. Unless otherwise stated, convex polygons are meant. Sides and angles of a polygon may be reckoned either clockwise or counter-clock- wise. 14. Prove that the sum of the inner angles of an ;«-side is {n — 2) straight angles. What is the sum of the outer angles ? 15. Find the angle in a regular {i.e. equiangular and equilateral) 3-side, 4-side, 5-side, 8-side, 12-side. (For proof that there is a regular «-side, see Art. 137.) 16. Show that a (convex) polygon cannot have more than three obtuse outer angles, nor more than three acute inner angles. 1 7. Two angles of a A are a and ^ ; find the angles at the intersection of their mid-rays. 18. If two A have their sides parallel or perpendicular in pairs, then the A are mutually equiangular. 19. The medial to the hypotenuse of a right A cuts the A into two isosceles A. 20. An angle in a A is obtuse, right, or acute, according as the medial to the opposite side is less than, equal to, or greater than, half the opposite side. EXERCISES I. 73 21. A medial will be greater than, equal to, or less than, half the side it bisects, according as the opposite angle is acute, right, or obtuse. 22. HP and Q be on the mid-normal of AB^ then AAFQ=i/SBFQ {= indicates congruence). 23. AB is the base, C the opposite vertex of an isosce- les A; show that ABN=BAM (i) when AM and BN are altitudes, (2) when they are medials, (3) when they are mid-rays of angles A and B, (4) when MN is normal to the mid-normal of AB. 24. F is any point within the A ABC ; show that AF+BF i {AB+BC-\- CA) . 25. ABC"' L and AB^C-"L are two convex polygons, not crossing each other, between the same pairs of points, A and L ; which is the longer ? Give proof. 26. /* is a point within AABC ; show that angle AFB > ACB and sum of angles sit F= 2{A + B -\- C). 27. F is equidistant from A, B, and C; show that angle AFB =2 {Single ACB). 28. Conversely, if angle ^P^ = 2 (angle ACB), angle BFC= 2 (angle BAC), and angle CFA = 2 (angle CBA), then F is equidistant from A, B, C. 29. The mid-rays of the angles at the ends of the trans- verse axis of a kite cut the sides in the vertices of an anti-parallelogram (Art. 99). 30. The four joins of the consecutive mid-points of the sides of a 4-side form a parallelogram. 31. The joins of the mid-points of the pairs of opposite sides and of the pairs of diagonals of a 4-side concur, bisect- ing each other. 74 GEOMETRY. 32. The mid-parallels to the sides of a A cut it into 4 congruent A. 33. What figures are formed by the mid-parallels when the A is right? isosceles? regular? 34. A parallelogram is a rhombus if a diagonal bisects one of its angles. 35. A parallelogram is a square if its diagonals are equal and one bisects an angle of the parallelogram. 36. From any point in the base of an isosceles A parallels are drawn to the sides ; the parallelogram so formed has a constant perimeter ( = measure round = sum of sides) . 37. The sum of the distances of any point on the base of an isosceles A from the sides is constant. 38. The sum of the distances of any point within a regular A from the sides is constant. — What if the point be without the A? 39. jP is on a mid-ray of the angle A in the A ABC', compare the difference of FB and PC : when P is within the A, and when P is without. 40. The inner mid-ray of one angle of a A and the outer mid- ray of another form an angle that is half the third angle of the A. 41. 6> is the orthocentre of the A ABC; express the angles A OB, BOC, CO A, through the angles A, B, C. 42. Do the like for the circum-centre 6* and the in-centre /. 43. The medial to the hypotenuse of a right A equals one-half of that hypotenuse. 44. The mid-rays of two adjacent angles of a parallelogram are normal to each other. EXERCISES I. IS 45. In a 5 -pointed star the sum of the angles at the points is a straight angle. What is the sum in a 7 -pointed star? 46. Parallels are drawn to the sides of a regular A, tri- secting the sides ; what figures result ? 47. A side of a A is cut into 8 equal parts, through each section point parallels are drawn to the other sides ; how are the other sides cut and what figures result? 48. Two A are congruent when they have two mid-tracts of two corresponding angles equal, and besides have equal (i) these angles and a pair of the including sides ; or (2) two pairs of corresponding angles ; or (3) one pair of corresponding angles and the correspond- ing angles of the mid-tract with the opposite side ; or (4) one pair of including sides and the adjacent segment of the opposite side. 49. Two A are congruent when they have two corre- sponding sides and their medials equal, and besides have equal (i) another pair of sides ; or (2) the angles of the medial with its side (in pairs) ; or (3) a pair of angles of the bisected side with another side, the angles of the medial with this side being both acute or both obtuse ; or (4) a pair of angles of the medial with an including side, the corresponding angles of the medial with its side being both acute or both obtuse. 50. Two A are congruent when they have a pair of cor- responding altitudes equal, and besides have equal 76 GEOMETRY. (i) the pair of bases and a pair of adjacent angles ; or (2) the pair of bases and another pair of sides ; or (3) the pairs of angles of the altitude with the sides ; or (4) two pairs of corresponding angles ; or (5) the two pairs of sides, when the altitudes lie both between or both not between the sides of the A. SYMMETRY. 90. We have seen that congruent figures are alike in size and shape, different only in place, and may be made to fit point for point, line for line, angle for angle. The parts that fit one on the other are said to correspond or be corre- spondent. Plainly only hke can correspond to like, as point to point, etc. Def. The ray through two points we may call the join of those points, and the point on two rays the join of the rays. 91. It is now plain that if ^ corresponds to A and B to B\ then the join of A and B must correspond to the join of A and B^ ; for in fitting A on A' and B on B' the ray AB must fit on the ray AB^ (why?). Also if the ray Z corresponds to L\ and M to M', then the join of L and M must correspond to the join of Z' and M' (why?). These facts are very simple but very important. We shall think of the plane as a thin double film, the one figure drawn in the upper layer, the other in the lower. 92. Two congruent figures may be placed anywhere and any way in the plane, but there are two positions especially important : ( i ) the one in which the one figure may be superimposed on the other by turning the one half of the plane through a straight angle about a ray called an axis ; (2) the one in which the one figure may be fitted on the SYMMETRY. 77 other by turning the one half of the plane through a straight angle about a point called a centre. Congruent figures in either of these two positions are called symmetric : in the first case axally, as to the axis of symmetry; in the second case centrally, as to the centre of symmetry. 93. In two symmetries, corresponding angles, like all other correspondents, are of course congruent ; but they are reckoned oppositely if the symmetry be axal, similarly if Fig. 61. 78 GEOMETRY. it be central. To parallels correspond parallels ; to normals, normals ; to mid-points, mid-points ; to mid-rays, mid-rays ; to the axis corresponds the axis, each point to itself; to the centre corresponds the centre itself (Fig. 6i). Elements, whether points or lines, that correspond to themselves may be called self-correspondent or double. It is also manifest that centre and axis are the only self- correspondents ; hence if a point be self-correspondent, it must lie on the axis in axal symmetry, or be the centre in central symmetry ; and if two counter half-rays be corre- spondent, they (or the ray) must be normal to the axis in axal symmetry, or go through the centre in central sym- metry. 94. These facts are all perfectly obvious, but a more vivid exemplification of the nature of these two kinds of symmetry may perhaps be found in the following : Suppose the axis of symmetry to be a perfect plane mirror; then either half of the plane may be treated as the reflection or exact image of the other, and will be the sym- metric of the other as to the mirror-axis. For the image of any point A is the point A such that the axis is the mid- normal of AA\ as we know from Physics ; also, on folding over the one half of the plane about the axis upon the other half, the point A falls on A^ (why?) ; hence A^ is the sym- metric of A as to the axis. Suppose the centre of symmetry 6" to be also a reflector ; then the reflection or image of any point A will be a point A such that S is the mid-point of the tract AA\ and on rotation through a straight angle about ^ the point A falls on A\ and the half-ray SA fits on the half- ray SA\ Hence either of two centrally symmetric figures is the exact image of the other reflected from the centre of symmetry S. SYMMETRY. 79 Note carefully that these two species of symmetry depend upon the two fundamental definitive properties of the plane : central symmetry upon the homoeoidality of the plane, axal symmetry upon the reversibility of the plane. Moreover, axally symmetric figures can not be fitted on each other without reversion, folding over ; by movement in the plane their corresponding parts can at best be ", and consider the 4-side DAD'A\ It is com- posed of two A, ADD^ and ADD\ symmetric with each other as to the axis X, and opposed along that axis. Hence the 4-side is itself symmetrical as to X Def. Such a 4-side, with an axis of symmetry, is called a kite. If we hold D fast, and let Z)' glide along X, the 4-side ADA^D^ remains a kite. We see that there are two kinds of kites, the convex kite, as ADA^D\ and the re-entrant, as ADAD^\ As the gliding point passes through iV^ the kite changes from one kind to the other, passing through the intermediate form of the symmetrical A. When the gliding point reaches a position Z>' such that ND = ND\ then the four sides of the kite are all equal (why?), and the kite becomes a rhombus (why?). In this case D and D^ are symmetric as to AA^ as an axis of sym- 84 GEOMETRY. metry. Hence the rhombus has two axes of symmetry ; namely, its two diagonals. In all cases the diagonals, A A and DD\ of the kite are normal to each other (why?). 100. Now consider a pair of points, B and B\ symmetric as to the axis X (Fig. 64). Then Xis mid-normal oi BB\ Fig. 64. If C and C be any other pair of symmetric points, then X is also mid-normal of CC ; hence BB^ and CC are parallel (why?). Also the tracts BC and B^ C are symmetric as to X (why?), and the 4-side BB^CC is itself symmetric as to the axis X, Hence the angles at C and C are equal. SYMMETRY. 85 also the angles at B and B^ are equal (why?) ; hence the angles at B and C and at B^ and C are supplemental (why?), and the 4-side BB'C'C is an anti-parallelogram (why ?) . Hence we see that another symmetric ^-side is an anti-parallelogram. It is plain that every anti- parallelogram is symmetric, for we know that the obHque sides prolonged yield an isosceles A. Let the student complete the proof. loi. There is only one kind of symmetric A, the isosceles. For, let^^^' (Fig. 65) be symmetric and A' correspondent /\ Fig. 65. to A. Then B must correspond to itself (why?) ; hence B must lie on the axis (why?) ; hence BA = BA^ (why?). Now let the student prove that (i) In a symmetric A the axis of symmetry is a medial ; (2) it is also a 7nid-ray ; (3) it is also a mid-normal. 86 GEOMETRY. Conversely^ let him show that A medial that is a mid-ray, or a mid-normal, is an axis of symmetry. 102. There are only two axally symmetric 4-sides ; namely, the kite and the anti- parallelogram. For, in a symmetric 4-side a vertex must correspond to a vertex (why?). Also, not all vertices can be on the axis (why?). Also, a vertex on the axis is a double point (why?). Also, the vertices not on the axis must appear in pairs (why?) ; hence there must be either two or four of them. If there be two only, then the other two are on the axis and the 4-side is a kite ; if there be four of them, we have just seen that the 4-side is an anti-parallelogram. 103. Now let us turn to the reciprocals. The reciprocals of the two points A and A^ symmetric as to the axis X will be two rays L, L\ symmetric as to the centre S. But rays symmetric as to a centre are parallel (why?) ; hence we have two parallels symmetric as to S, which is midway between them. The rays are symmetric as to any other point S' mid- way between them (why?). The piece of plane between these parallels is called a parallel strip, or band (Fig. 66). ^'-^ M •s' L Fig. 66. SYMMETRY. 87 But what corresponds to the point D on the axis X} The answer is : a ray R through 6" (why?). Hence to the sym- metric A of the three points A, A\ D, there corresponds the figure formed by two parallels L, L\ and a transverse R through 5, — a so-called half-strip. This is truly a three- side, but not apparently a A (3 -angle), for the parallels do not meet in finity, in regions accessible to our experience. Hence, instead of saying that the reciprocal of a A in axal symmetry is a A (3-angle or 3-point) in central sym- metry, we should have said, accurately, that the recipro- cal of a A in axal symmetry is a 3-side (or trilateral) in central symmetry, which will always be a A except when sides are parallel or all concur. In higher Geometry it is very convenient to remove this apparent exception by using this form of expression : the parallels meet not in finity, but in infinity. 104. It is indeed plain that A A can have no centre of symmetry. For, since vertex corresponds to vertex, and since corre- spondents appear in pairs, one vertex must be a double point ; hence it would have to be the centre S (why ?) . But the other two vertices would have to lie on a ray through S, being correspondents ; hence the three vertices would be collinear, and the A would be flattened out to a triply-laid ray. 105. But there is a centrally symmetrical 4-side ; namely, the parallelogram. For, consider once more the kite AXA^X and let us reciprocate it into a centrally symmetric figure (Fig. 67). To the axis XX^ will correspond the cen- tre S; to the symmetric pair of rays ^Jfand y4'X will corre- spond a symmetric pair of points P and P ; to the join of those on the axis X will correspond the join of these through A'^ 88 GEOMETRY, the centre {FF^y. Similarly, to the symmetric rays AX' and A'X' will correspond the symmetric points Q and Q\ and to the join X' will correspond the join QQ. Also, AX and AX' have a join A while A'X and ^'X' have a join ^', and these joins are symmetric as to the axis XX' ; recipro- FlG. 67. cally, /'and Q have a join Z*^, and F' and ^' have a join FQ'y and these joins are symmetric as to ^S; that is, they are parallel (why?). Similarly, FQ' and F' Q correspond to B {AX, A'X') and B' (A'X, AX') ; but B and B' are sym- metric as to XX' (why?) ; hence FQ' and F'Qave symmetric as to S, i.e. are parallel. Hence FQ FQ is symmetric as to S, and is 2, parallelograin. q. e. d. 106. We may indeed see at once that since any two par- allels are centrally symmetrical as to any mid-point, a pair SYMMETRY. 89 of parallels or a parallelogram is symmetric as to the common mid-way point, the intersection of the diagonals. But the foregoing reciprocation is instructive, as illustrating in detail the method to be pursued, and as showing the intimate rela- tion of the different symmetric quadrilaterals ; namely, the parallelogram is the common reciprocal of doth '^tQ and axiti- parallelogram, which are thus seen to be really one. 107. Central symmetry does not in general imply any- thing at all with respect to axal symmetry in a figure. We may draw through any point S any number of rays and lay off on each from -S a pair of counter tracts SF and SP\ SQ and SQ', etc. No matter how FQ, etc., be chosen, the figure so obtained will be centrally symmetric as to 6"; but it may have no axal symmetry whatever. Neither does axal symmetry in general imply any central symmetry, but we may estabhsh the following important Theorem. — Any figure with two rectangular axes of symmetry has also a centre of symr?ietty ; namely, the inter- section of those axes. p V # p '^^ ^.--^ ^^ ^^ \ 3'' P Y Fig. 68. 90 GEOMETRY. Data : XX^ and FF' two rectangular axes, P any point of a figure symmetric as to these axes (Fig. 68). Proof. The point P^ symmetric with P as to XX^ is a point of the figure (why?) ; also P" symmetric with P^ as to yy is a point of the figure (why?) ; so too is /^'" (why?) ; the figure pp'P^p^'^ is a rectangle (why?), its diagonals halve each other, and SP= SP" = SP' = SP'". Hence ^ is a centre of symmetry, q. e. d. THE CIRCLE. 1 08. We have already discovered the existence of a homoeoidal plane curve not reversible and have named it circle. Defs. A ray cutting a curve is called a secant, as L ; the part of the secant intercepted by the curve, or the tract between two points of the curve, is called a chord, as AB. A finite part of a curve is called an arc. A chord and an arc with the same two ends are said to subtend each other. Also, the intercept of any line between the ends of an angle is said to sub fern/ the angle. Thus ^Cand Z>£ subtend the angle O (Fig. 69). Fig. 69. Th. XLIX.] THE CIRCLE. 91 109. Theorem XLVIII. — Congruent arcs subtend con- gruent chords. Proof. Let the arcs AB and CD be congruent ; then we may fit ^ on C and at the same time B on D ; then the chords AB and CB fit throughout (why ?) . q. e. d. N.B. We can not convert this proposition at once (why?) (Fig, 70). Fig. 70. no. Theorem XLIX. — A closed curve is cut by a ray in an even number of points (Fig. 71). Proof. Let Z be a ray, C any closed curve. Suppose a point P to trace out the ray Z. At first P is without the curve, at last it is also without the curve ; hence P has crossed the curve going out as often as it has crossed the curve going in, for every entrance there is an exit ; hence the points of intersection appear in pairs, their number is even, as o, 2, 4, 6, ... 2 n, q. e. d. 92 GEOMETRY. [Th. L. These preliminary or auxiliary theorems, which prepare the way for a theorem to follow, are sometimes called lemmas (XyjfjifjLa = assumption, premise, support, prop) . *iii. Theorem L. — A circle has an axis of symmetry through every one of its points (Fig. 72). Fig. 72. Proof. Let D be any point of a circle. Take any arc DP, and slip it round till P falls on D and D on P' ; this is possible (why ?) . Then PDP' is a symmetrical A (why ?) ; and its axis of symmetry DR halves normally the chord PP\ and also halves the angle PDP' (why?). Now take any other arc DQ and slip it round till Q falls on D and D on Q\ so that P)Q and QD are congruent. Then the chords DQ and DQ are congruent (why?). Also, on taking away the congruents DP and DP^ we have left PQ and P^Q as congruent remainders. Hence the chords PQ Th. LI.] THE CIRCLE. 93 and PQ are congruent (why?). Hence the A PDQ and PDQ are congruent (why?) ; hence the angles PDQ and PDQ are equal (why?) ; hence DR halves also the angle QDQ (why?). But the A QDQ is symmetric (why?) ; hence DR is also its axi? of symmetry, and Q and Q are symmetric points of the circle ; hence any point of the circle has its symmetric point as to DR; i.e. DR is an axis of symmetry of the circle. Moreover, D was any point of the circle ; hence through any point of the circle passes an axis of symmetry, q. e. d. Def. A ray halving a system of parallel chords is called a diameter; the chords and diameter are called conjugate to each other. Corollary i. In a circle a diameter is normal to its con- jugate chords. Corollary 2. Every mid-normal to a chord in a circle is a diameter and halves the subtended arcs. *ii2. Theorem LI. — A circle has a centre of symmetry (Fig. 72). For the ray through D must cut the circle in some second point, as R (why?), and as the ray turns round from the position DR to the reversed position RD, through a straight angle, it must pass through some position, QQ, normal to its original position (why ?) . Hence for any axis of symmetry there is another normal thereto and their intersection is a centre of symmetry (why ?) . q. e. d. N.B. There is only one centre of symmetry (why?). Def. This centre of symmetry is named centre of the circle. It is often convenient to call the whole ray through the centre a centre ray or line^ and to restrict the term diameter to the centre chord. 94 GEOMETRY. [Th. LII. Corollary i. All diameters go through the centre, and halve each other there ; conversely, chords halving each other are diameters. Def. Two diameters each halving all the chords parallel to the other are called conjugate. Corollary 2. In the circle two diameters normal to each other are conjugate ; and conversely, two conjugate diameters are normal to each other. N.B. Other curves, as Ellipse and Hyperbola, have conjugate diameters not in general normal to each other (Fig. 73)- *ii3. Theorem LII. (Fig. 74). All diameters of a circle are equal Fig. 73. Fig. 74. Proof. Let DR and D^R^ be two diameters. The figure DD'RR is a parallelogram (why?), and DV is parallel to RR^ ; hence the mid-normal of these parallels is a diameter Th. LIL] the circle. 95 through the centre S ; hence SD and SD' are symmetric and equal ; hence DR = D'R\ Q. e. d. JDef. A half-diameter, from centre to circle, is called a radius. Corollary i. All radii of a circle are equal ; or, all points of a circle are equidistant from the centre. Corollary 2. Every parallelogram inscribed in a circle is a rectangle. N.B. By help of this important property the circle is commonly defined as a plane curve all points of which are equidistant fro7n a point within called the centre. The com- mon distance of all points of the circle from the centre is often called the radius. We have deduced this property from the homoeoidality ; conversely ^ we may deduce the homoeoidality from this property taken as definition. But if there were no such surface as the plane, at least for our intuition, the circle might still exist on the sphere-surface, without centre, but with the body of its properties unimpaired. Hence it seems better to define the circle by its intrinsic homoeoidality than by its extrinsic centraHty. Corollary i. All points within the circle are less, and all points without are more, than the radius distant from the centre. Defs. The two symmetric halves into which a diameter cuts a circle are called semicircles. The part of the plane bounded by an arc and its chord is called a segment ; the part bounded by an arc and the two radii to its ends is called a sector. If the sum of two arcs be a circle, we may call them explemental, the one minor ^ the other major ; every chord belongs equally to eacli of two explemental arcs, but in general, unless otherwise stated, it is the minor that 96 GEOMETRY. [Th. LIII. is referred to. Two arcs whose sum is a half-circle are called supplemental ; two whose sum is a quarter- circle or quadrant are called complemental. Corollary 2. All circles of the same radius are congruent ; also, all semicircles of the same radius are congruent, and all quadrants of the same radius are congruent. Corollary 3. Any circle may be slipped round at will upon itself about its centre as a pivot, Hke a wheel about its axle, without changing in the least the position of the whole circle. 114. From the foregoing it is clear that if we hold one point of a ray fixed, and turn the ray in the plane about the fixed point, every other point of it will trace out a circle about the fixed point as a centre. An instrument, one point of which may be fixed while the other is movable about in a plane, is called a compass or pair of compasses, and is both the simplest and the most important of all instruments for drawing. 115. Theorem LIII. — Through any three points not col- linear one, and only one, circle may be drawn. Proof. Let A, B, C be the three points not collinear (Fig. 75). We have already seen that the mid-normals to the tracts AB, BC, CA concur in a point 6* equidistant from A, B, and C; hence a circle about »S with radius d passes through A, B, C. Also there is only one point thus equi- distant from A, B, C (why?) ; hence there is only one circle through A, B, C. Q. e. d. Def. The circle through the vertices A, B, C, of a A is called the circum-circle of the A. Corollary i. A A, or a triplet of points, or a triplet of rays, determines one, and only one, circle. Th. LIIL] THE CIRCLE. 97 Corollary 2. Through two points, A and B, any number of circles may be drawn. Their centres all lie on the mid- normal of AB. Corollary 3. As BC turns clockwise about ^ as a pivot, the intersection S, the centre of the circle through A, B, C, retires upward ever faster and faster along the mid-normal N of AB ; when C becomes collinear with A and B, the inter- FiG. 75. section of the raid-normals of AB and BC vanishes from finity, or retires to infinity, as the phrase is. As BC keeps on turning, S reappears in finity below and moves slower and slower upward along the mid-normal. Moreover, a circle passes through A, B, and C, no matter how close C may lie to the ray AB, nor on which side of it : only as C falls upon the ray does the centre of the circle vanish into infinity ; that is, we may draw a circle that shall fit as close to the ray AB as we please, though not upon it, by retiring the centre far 98 GEOMETRY. [Th. LIV. enough. Hence a ray may be conceived as a circle with centre retired to infinity ; it is strictly the limit of a circle whose centre has retired, along a normal to it, without limit. ii6. Theorem LIV. — A circle can cut a ray in only two points. For there are only two points on a ray at a given distance from a fixed point (why?), q. e. d. 117. Theorem LV. — Secants that make equal angles with the centre ray (or axis) through their intersection intercept equal arcs on the circle. Proof. For both the two semicircles and the two secants are symmetric as to the axis IS (why?) ; hence, on folding over the one half-plane upon the other, A falls on A\ B on B\ arc a fits on arc a\ and chord c on chord c^ (Fig. 76). Q. E. D. Th. LVI.] the circle. 99 Conversely, Secants that intercept equal arcs make equal angles with the axis through their intersection. Proof. Let L and Z' intersect equal arcs AB and AB\ Draw the mid-normal of A A ; it is an axis of symmetry (why?). On folding over the left half- plane upon the right half-plane, A falls on A^ and B onB' (why?) ; hence AB and A'B' are symmetric ; hence they meet on the axis and make equal angles with it (why?), q. e. d. Corollary i. Equal chords are equidistant from the cen- tre ; and conversely , Chords equidistant from the centre are equal. Corollary 2. The greater of two unequal chords is less distant from the centre. Corollary 3. A diameter is the greatest chord. Corollary 4. Arcs intercepted by two parallel chords are equal. Corollary 5. Equal chords or arcs subtend equal central angles (angles at the centre), and conversely. Corollary 6. Of two unequal chords or arcs, the greater subtends the greater central angle. What figure is determined by two parallel chords and the chords of the intercepted arcs ? By two secants that inter- cept equal arcs and the central normals thereto ? 118. Theorem LVI. — A central angle sub te tided by a cer- tain arc (or chord) is double the peripheral angle subtended by the same (or an equal) arc (or chord) (Fig. 77). Proof. Let ASB be a central angle, and APB be a peripheral angle (periphery = circumference, the circle itself), subtended by the same arc or chord AB. Draw the 100 GEOMETRY. * [Th. LVI. diameter PZ>. Then the A ASP and BSF are isosceles (why?) ; hence the angle ASD = 2 angle AFD, and angle BSD = 2 angle BPD (why ?) ; hence angle ASB = 2 angle APB. Q.E.D. P ^ Fig. 77. Corollary i . All peripheral angles subtended by (or stand- ing on) the same or equal chords or arcs are equal. Hence, as P moves round from A to B, the angle APB remains unchanged in size. Def. An angle with its vertex on a certain arc, and its arms passing through the ends of that arc, is said to be inscribed in that arc. Hence for an angle to be inscribed in a certain arc, and for it to stand on the explemental arc, are equivalent. Corollary 2. All angles inscribed in the same or equal arcs of the same or equal circles are equal. Corollary 3. As the vertex Z' of a peripheral angle sub- tended by an arc (or chord) AB, in passing round a circle goes through either end of the arc (or chord), the angle itself leaps in value, changes to its supplement. Th. LVIIL] THE CIRCLE. 101 119. Theorem LVII. — The locus of the vertex of a given angle standing on a given tract is two synu?ietric circular arcs through the ends of the tract (Fig. 78). \0 Fig. 78. Proof. Let F be the vertex of the given angle, in any position, standing on the tract AB. Through A, F, and B draw a circular arc subtended by AB. We have just seen that as long as F stays on this arc, the angle F remains the same in size. Moreover, the point F cannot be without the arc, as at O, because the angle A OB is less than AFB (why?) ; neither can it come within the arc, as to /, because the angle AIB is greater than AFB (why ?) ; hence so long as the angle is constant in size the vertex must remain on the arc AFB or on its symmetric arc AF^B^ of which plainly the same may be said. q. e. d. 120. Theorem LVIII. — The angle inscribed in a semi- circle (or standing on a semicircle or diameter) is a right- angle (Fig. 79). 102 GEOMETRY. [Th. LVIII. Proof. Let ^^C be any angle in a semicircle. Then it is half of the central angle ASC (why?), which is a straight angle (why?), q. e. d. Fig. 79. Now let the vertex B, the intersection of the rays L and N, move round the circle toward C ; the angle ABC re- mains a right angle, no matter how close B approaches to C ; moreover, when B passes C, into the lower semicircle, the angle remains a right angle (why?). That is, the angle 2itB remains a right angle, no matter from which side nor how close B approaches to C. Hence it is a right angle even when B falls on C, But then the ray L falls on the diame- ter A C, hence the ray N takes the position T normal to the diameter (or radius) at its end. Such a normal to a radius at its end is called a tangent to the circle at the point of tangence (or touch or contact^ C. Def. A ray normal to a tangent to'a curve at the point of touch is called normal to the curve itself. Hence Corollary. All radii of a circle are normal to the circle ; and conversely, all normals to a circle are radii of the circle. Th. LX.] THE CIRCLE. 103 121. Theorem LIX. — All points on a tangent^ except the point of contact, lie outside of the circle. Proof. For the point of touch is distant radius from the centre (why?), and all other points, as Z>, of the tangent are further from the centre (why?) ; hence all other points of the tangent are without the circle (why?), q. e. d. 122. Theorem LX. — The point of tangence is a double point. Proof. For it is on a diameter, or axis of symmetry, of the circle, and every such point is a double point with respect to that axis. Independently of this consideration, it is seen that the chord CB becomes the tangent CT when, and only when, the points B and C fall together in C. Fig. 8o. Still otherwise, let AB be any chord of a circle about (Fig. 8o) O. Draw the mid-normal OD. Now let the circle shrink about the centre O : the points A and B move 104 GEOMETRY. [Th. LXI. towards each other, and as D is always mid-way between them they finally fall together in D, and their join is tan- gent at D to the circle of radius OD. Def. Two points thus falUng together in a double point are called consecutive points. Accordingly we may define a tangent to a circle (or to any curve) as a ray through two consecutive points of the circle (or curve). Adopting this definition, let the student prove 123. Theorem LXI. Every tangent to a circle is normal to a radius at its end ; conversely, Every normal to a radius at its end is tangent to the circle. 124. Theorem LXII. The angle between a tangent and a chord equals the peripheral angle on the same chords or equals half the angle of the chord (Fig. 81). Proof. For if DT be a diameter, then the angles BDT and BTA are equal, being complements of the same angle BTD (why?). Or thus : TB 've> d. chord, and TA is also a Th. LXIV.] THE CIRCLE. 105 chord, through the double point T ; hence the angle BTA is a peripheral angle standing on the arc TB. q. e. d. 125. Theorem LXIII. — The angle between two secants is half the sum or half the difference of the angles of the intercepted arcs, according as the secants intersect within or without the circle. Proof. For on drawing AB^ the angle / is seen (Fig. 82) to be the sum, and the angle O the difference, of the Fig. 82. angles at A and B^ standing on the arcs AA' and BB\ Q. E. D. 126. Theorem LXIV. — An encyclic quadrangle has its opposite angles supplemental. Proof. For the angles B and D are halves of the two central angles ASC and CSA, whose sum is a round angle. Hence the sum of B and Z> is a straight angle, q. e. d. 106 GEOMETRY. [Th. LXV. 127. Theorem LXV. — Conversely, A quadrangle with its opposite angles supplemental is encyclic (Fig. ^2)). Fig. 83. Proof. Let ABCD be the quadrangle with the angles A and C, ^ and D, supplemental. About the A ABC draw a circle. If P be any point on the arc of this circle exple- mental to ABC, then the angle APC xs^ the supplement of ABC \ but if P be not on this arc, then the angle A PC is either greater or less than that supplement (why?). Now the angle D is that supplement ; hence D is on the arc. Q. E. D. 128. Relations of circles to each other. Suppose two circles K and K^ of radii r and r' to be concentric, i.e. to have the same centre O. Then, plainly, the distance between them measured on any half-axis OR is r—r\ the difference of the radii. Draw tangents AT, A^T', where 00^ cuts the circles. They are parallel (why?). Now let the centre of K' move out on 6>(9' a distance r — /; then ^ falls on A^ and A'T^ on. AT; the circles have a common tangent at A and are said to touch each other innerly at ^ (Fig. 84). Th. LXVIL] THE CIRCLE. 107 Now let (9' move still further along 00^ ; then the circles will lie partly within, partly without, each other ; they will intersect at two points, and only two (why?), symmetric as to 00^ (why?), namely /'and P ; hence "Tk' i<' yfe) Fig. 84. Theorem LXVI. — The cotnmon axis of two circles is the mid-7iormal of their conwion chord. When O^ is distant r + / from O, the circles He without each other, but still have a common tangent (why ?) and are said to touch outerly. As O^ moves still further away from O, the circles cease to touch and henceforth He entirely without each other. Thus we find that there are three critical positions depend- ing on the distance d between the centres O and O^ : d=o, when the circles are concentric. d=z r— r\ when the circles touch innerly. d= r-\- r', when the circles touch outerly. There are also three intermediate positions : For o < d < r—r^ the one circle is withi?i the other. For r—r^<. d and C are equal ; hence A' = A'' ( why ?) q. e. d. Fig. 89. *i34. Theorem LXXII. — T/ie mid-points of the sides of a A, the feet of its altitudes, and the 7tiid-points between its 07-thocentre and vertices^ are nine encyclic points. Proof. Let a circle through X, V, Z, the mid-points of the sides, cut the sides in three other points, O] V, W. Then the angle ZXY= angle A (why?), and also = angle ZVY (why ?) ; therefore the A AZV is symmetrical. Hence the A ZVB is also symmetrical, Z is equidistant from A, V, and By and the angle AVB is a right angle (why ?) ; so also the angles at C^and IV; i.e. the circle through the mid-points of the sides goes through the feet of the altitudes (Fig. 90). Again, if the circle cuts the altitudes at P, Q, R, then the angle F/'^= angle VZIV (why?) = 2 angle F^^(why?). Moreover, A, F, O, IV, are encyclic (why?) ; hence AO is a diameter of the circle through them (why?) ; and VAW is a peripheral angle standing on the arc VIV; hence the 112 GEOMETRY. [Th. LXXIII. double angle VjRJVmust be the central angle of the same arc ; t.e. P is the mi^-point between a vertex and orthocen- tre : so, also, are Q aftd R, similarly, q. e. d. Def. This remarkable circle is called the 9-point circle, or circle of Feuerbach, of the A ABC, Corollary. The radius of the 9-point circle is half the radius of the circumcircle. 135- Def. A Polygon all of whose sides touch a circle is said to be circumscribed about it, and the circle is said to be inscribed in the polygon. Theorem LXXIII. — A circle may be inscribed in any A. Proof. Let ABC be any A (see Fig. 59). Draw the inner mid-rays of the angles at A, B, C ; they concur in the in-centre /of the A, equidistant from the three sides (why?). About this point as centre with this common distance as radius draw a circle ; it will touch the three sides of the A (why and where ?) . q. e. d. N.B. We have seen that the outer mid-rays of the angles concur in pairs with the inner mid-rays of the angles in the three ex-centres £1, £2, -^sl also equidistant from the sides Th. LXXIV.] THE CIRCLE. 113 (Fig. 60). The circles about these touch only two sides innerly, but the third side outerly, ai^ hence are called escribed, or ex-circles. • Corollary, Four, and only four, circles touch, each, all the sides of a A. 135 a. Theorem LXXIV. — In^ a 4-side circumscribed about a circle the sums of the two pairs of opposite sides are equal (Fig. 91). Fig. 91. Proof. The sum of the four sides is plainly 2/4-2/^+22^ -f- 2Z£/, and the sum of either pair of opposites is t-\-u-\-v-\-w. Q. E. D. Conversely, If the sums of two pairs of opposite sides of a 4-side be equals the 4-side is circumscribed about a circle. Proof. Let two counter sides, AB and DC meet in /, and inscribe a circle K in the triangle ADL Through B 114 GEOMETRY. [Th. LXXV draw a tangent (Fig. 92) to A' at U, and let it cut DI dX C Then since ABCD is circumscribed about K, we have or Fig. 92. AB-\-CD = BC-\-DA. Also AB+CD =BC + DA (why ?). Whence CD-CD^BC-B C\ CC = BC-BC. Hence Cand C fall together (why? Art. 56). q. e. d. 136. Theorem LXXV. — The tangent-length fro?n a ver- tex of a /\ to the in-circle equals half the perimeter of the A less the opposite side (Fig. 93). Th. LXXVI.] THE CIRCLE. 115 Proof. For the sum of CE ^ CD -\- BD + BF is plainly 2a (why?) ; subtract this from the whole perimeter, a -{- b -\- c^ and there remains AE -\- AF— a -\-b -\- c — 2a^ or AE = b -\- c— a . E, —^ = AF. Q. E. D. FIG. 93. ; It is common and convenient to denote the perimeter (Fig. 93) (= measure round = sum of sides) by 2J-; then we see that the tangent-lengths from A^ B, C, are s — a^ s — bj s — c. Corollary. The tangent-length from any vertex, A, of a A to the opposite ex-circle and the two adjacent ex-circles are s, s—b, s — c. Hence s — a, s, s — b, s — c, are the four tangent-lengths from any vertex, ^, of a A to the in- circle and the three ex-circles. These relations are useful and important. 137. Theorem LXXVI. — There is a regular n-side. Proof. For the angle is a continuous magnitude (why?) ; hence there are angles of all sizes from zero to a round 116 GEOMETRY. [Th. LXXVI. angle ; hence there is an angle, the - part of a round angle, n such that, taken n times in addition, the sum will be a round angle. Suppose such an angle drawn, whether or not we can actually draw it, and suppose n such angles placed con- secutively around any point O, so as to make a round angle. In other words, suppose n half-rays drawn cutting the round angle about O into n equal angles. Draw a circle about O, with (Fig. 94) any radius, and draw the 7? chords Fig. 94. subtending the n equal central angles. These chords are all equal (why?), and subtend equal arcs, and they form an ;/-side. Moreover, the angle between two consecutive sides Th. lxxviil] the circle. 117 is constant in size, because it stands on the ^^ ~ ^ part of tlie circle. Hence the «-side is both equilateral and equian- gular ; that is, it is regular, q. e. d. Corollary. The inner angle of a regular «-side is the part of a straight angle. \ (^") Find the value in degrees of the inner angles of the first ten regular «-sides. N.B. The foregoing demonstration merely ; settles the question of the existence or logical possibility of the regular ;/-side. The problem of actually drawing such a figure is one of the most intricate in all mathematics, and has been solved only for certain very special classes of values of n. But in order to discover the properties of the figure, it is by no means necessary to be able to draw it accurately. It is only since 1864 that we have known how to draw a straight line or ray exactly. 137 a.' Theorem LXXVII. — The vertices of a regular n-side are encyclic (Fig. 94). ftroof. Through any three vertices, as A, B, C, of a regular ;2-side, draw a circle K ; about C with radius CB draw another circle. The fourth vertex D must lie on this circle (why?). If it lie on the circle K, then the angle BCD — angle ABC, as is the case in the regular «-side. Neither can it lie off of K, as at Z>' or D", because then the angle BCD' or BCD'' would not equal angle BCD (why?), and hence would not equal angle ABC. Hence the next vertex must lie on the same circle K, and so on all around, q. e. d. 138. Theorem LXXVIII. — The sides of a regular n-side are pericyclic (that is, they all touch a circle). 118 GEOMETRY. [Th. LXXIX. Proof. For, on drawing the radii of the circumcircle K (Fig. 95) to the vertices, we get n congruent symmetric A A O Fig. 95. (why?). The altitudes of all are the same (why?) ; with this common altitude as radius draw another circle, K\ about the same centre. It will touch each of the sides (why?). Q. E. D. Corollary. The points of touch of the sides of the regular circumscribed /z-side are mid-points of the sides. 139. Theorem LXXIX. — The points of touch of a regular circumscribed n-side are the vertices of a regular inscribed n-side. Proof. Connect the points of touch consecutively. Then the A so formed are all congruent (why?) ; hence the joining chords are equal; hence the arcs are equal; hence the Theorem, q. e. d. CIRCLE AS ENVELOPE. 119 THE CIRCLE AS ENVELOPE. *i4oa. Thus far we have regarded the circle from various points of view ; from the most familiar it was seen to be the locus of a point in a plane at a fixed distance from a fixed point. An almost equally important conception of the curve treats it not as the locus of a point, but as the envelope of a ray. If the point P moves in the plane always equidistant from (9, then its locus is the circle, on which it may always be found ; also, if the ray R moves about in the plane always equidistant from 6>, then its envelope is the circle, on which it may always be found, on which it Hes, which it continually touches. The point traces the circle, the ray envelops the circle, which is accordingly called the envelope (i.e. the enveloped curve — French enveloppee) of the ray. In higher mathematics the notion of the ray, instead of the point, as the determining element in the nature of a curve, attains more and more significance. In this text we are confined to the circle — the envelope of a ray in a plane, at a fixed distance from a fixed point. *i4ob. It is not only rays, however, that may envelop a curve ; but circles, and in fact any other curves. Thus, let the student draw a system of equal circles, having their centres on another circle ; the envelope will at once be seen to be a pair of concentric circles. Let him also find the envelope of a system of circles equal and with centres on a given ray. In general, let him find the envelope of a circle whose centre moves on any given curve. Lastly, let him draw a large number of circles all of which pass through a fixed point, while their centres all lie on a fixed circle, and let him observe what curve they shadow forth as envelope. Show that as the pole of a chord (or ray) traces a circle. 120 GEOMETRY. the chord itself envelops a concentric circle, and con- versely. Show that tangents from two points on a centre ray form a kite, and conversely. Also the chords of contact are parallel, and conversely. O is the centre of a circle, P any point without it. Show how to find the point of touch of the tangents from P^ by drawing a circle about O through P and a tangent where OP cuts the given circle. CONSTRUCTIONS. 140. Hitherto, in our reasoning about concepts, figures have not been at all necessary, though exceedingly useful in making sharp and precise our imagination of the relations under consideration, in furnishing sensible examples of the highly general notions that we dealt with. The conclusions reached thus far all He wrapt up in axioms and in our defi- nitions of point, ray, and circle, and our work has been one of explication only ; we have merely brought them forth to light. Our demonstrations have not presumed ability to draw accurately, and would remain unshaken if we could not draw at all. Nevertheless, for many practical purposes, it is ex- tremely important and even indispensable that we actually make the constructions and draw the figures that thus far we have merely supposed made and drawn. 141. What is meant by drawing a ray, circle, or any line? Any mark, whether of ink or chalk, though a solid, may be treated as a line by abstraction. Only its length, not its width nor thickness, concerns us. How to make not just any mark, but some particular mark called for, is our prob- lem {Trpo/SXrjiJLa = anything thrown forward as a task), and CONSTRUCTIONS. 121 its solution consists accordingly of two parts, the logical and the mechanical. The first is accomphshed by fixing exactly in thought the position of all the geometric elements (points, rays, circles) in question ; the second, by making marks that by abstraction may be treated as these elements. Now, a point is fixed as the join of two rays, a ray as the join of two points (by what axiom?) ; a circle is fixed or determined by its centre and radius (why?), or by three points on it (why?). Accordingly, when we know two rays through a point, or two points on a ray, or centre and radius, or three points of a circle, we know the point, or ray, or circle completely. The logical part of our work is finished, then, when we determine every point as the join of two known rays, every ray as the join of two known points, every circle as drawn through three known points or about a known centre with a known radius. The mechanical part of the solution requires us to put and keep a point in motion along a circle or a ray. Circular motion is brought about by the compasses already described (Art. 114), of which the shape is arbitrary, the necessary parts being merely a fixed point rigidly connected in any way with a movable point. But in the ruler one edge is supposed made straight to begin with, so that a pencil point gliding along it may trace a straight mark. Hence the use of the ruler is really illogical, since it assumes the problem of draw- ing a ray or straight line as already solved in constructing the straight edge. To say that, in order to draw a straight line J we must take a straight edge and pass a pencil point along it, is no better logically than to say that, in order to draw a circular line, we must take a circular edge and pass a pencil point along it. The question at once arises. How make the edge straight or circular in the first place ? It was not until 1864 that PeaucelHer won, though he did not at once receive, the Montyon prize from the French Academy 122 GEOMETRY. by solving the thousand-year-old problem of imparting rec- tilinear motion to a point without guiding edge of any kind (Page ooo). But, though the ruler is logically valueless, it is practically invaluable, even after the great discovery of Peau- cellier. Its edge being assumed as straight and of any desired length, and a pair of compasses of adjustable size being given, we now make the following Postulates : I. About any point may be drawn a circle of any radius, II. Through any two points may be drawn a ray (more strictly, a tract of any required length) . Corollary. On any ray from any point on it we may lay off a tract of any required length. These are the only instruments used or postulates assumed in the constructions of Elementary Geometry. 142. The fundamental relations of rays to each other are two : Normality and Parallelism. Hence Problem I. — To draw a ray normal to a give ft ray. Since there are many rays normal to a given ray, to make the problem definite we insert the Hmiting condition, through a given point. Two cases then arise : A. When the given point is on the given ray. All we can do is to draw a circle about the point P. It cuts the ray at two points, A and A\ symmetric as to F. Hence the mid- normal of AA^ is the normal sought. Hence any point on this normal Hes on two circles of equal radius about A and A\ Hence (Fig. 96) Solution. From the given point P lay off on the given ray two equal tracts PA, PA'. About A and A' draw two equal circles. Through their points of intersection draw their common chord. It is the normal sought. CONSTR UCTIONS. 123 Proof. For it is the mid-normal of AA\ since it has two of its points equidistant from A and A\ and P is the mid- point of AA\ Fig. 96. Query : What radius shall we take for the circles about ^and^'? B. When the given point is not on the given ray. All we can do is to draw a circle about the given point P. Let it cut the ray at A and A\ Then the mid-normal to A A' is the normal required (why?). Hence (Fig. 97) Solution. Determine the points A, A^ on the ray by a circle about the given point P; then proceed as in the first case (A). 124 GEOMETRY. Proof. For the mid-normal of A A goes through P (why?). /K Fig; 97. Query : What radius shall \^^e take for the circle about P ? 143. Problem II. — To dmw a parallel to a given ray. Since there are many parallels to every ray, to make the problem definite we must insert the limiting condition, through a given point ; then it becomes perfectly definite (why?). Manifestly the point must be not on the ray (why?). We now reflect that a transversal makes equal corresponding angles with parallels, and we have just learned to draw a normal transversal. Hence (Fig. 98) Solution. Through the point draw a normal to the ray; through the same point draw a normal to this normal. It will be the parallel required. Proof. For it goes through the point and is parallel to the ray (why?). These two problems have been discussed at such length as being the hinges on which nearly all others turn. At CONSTRUCTIONS. 125 the end of a problem is sometimes written Q. E. f. = quod erat faciendum =1 which was to be done, and translates the Euclidean OTrcp cSet Trpa^ai. \/l TX ^ TX Fig. 98. 144. Problem III. — To bisect a given tract, or to draw the mid-normaJ to a given tract, AB, Proceed as in Problem I. 145. Problem IV. — To bisect a given angle. Solution. About the vertex draw any circle cutting the arms at A and A\ and draw the mid-normal of AA\ It is the mid-ray sought (why ?) . Corolla?'}'. Show how to bisect any circular arc AB. 146. Problem V. — To bisect the angle between tivo rays whose join is not given (Fig. 99). 126 GEOMETRY. We reflect that the join AA^ of two corresponding points on the rays makes equal angles with the two rays that form the angle. Hence Solution. From any point P oi L draw the normal to it, cutting M at Q. From Q draw the normal to M. Bisect Fig. 99. the angle at Q between these two normals by the mid-tract QR. Draw the mid-normal of QR. It is the mid-ray sought (why?). 147. Problem VI. — To multisect a given tract AB (Fig. 100). L B Fig. icxj. CONSTR UCTIONS. 127 Solution. Through either end of the tract, as A^ draw any ray, and lay off on it from A successively n equal tracts, Z being the end of the last. Draw BL. Through the ends of the equal tracts draw parallels to BL. They cut AB into n equal parts (why ?) . 148. Problem VII. — To draw an angle of given size, i.e. equal to a given angle (Fig. loi). , /' y I Solution. At any point A of either arm of the given angle O erect a normal to OA cutting the other arm at B. From any point O on any other ray lay off O^A' = OA, and normal to the ray erect A'B' = AB and draw O'B'. Then angle O' = angle O (why ?) . When does this construction fail ? How proceed then ? 149. Problem VIII. — To draw a tract of given length subtending a given angle and parallel to a given ray. Data : O the given angle, L the ray, a the length (Fig. 102). Solution. Through any point P of either arm of the angle draw a parallel to the ray, and lay off on it towards the other 128 GEOMETRY. arm a tract PA of the given length a. Through A draw a parallel to OP, cutting the other angle arm at Q ; through Q draw a parallel to PA meeting OP2X R. QR is the subtense sought (why?). Fig. 102. 150. Problem IX. — To construct a A : A. When alternate parts (three sides or three angles) are given. Solution. About the ends of one side AB, with the other sides for radii, draw circles meeting in C. Then ABC is the A sought (why?) (Fig. 103). Fig. 103. How many such A may be drawn on the same base AB'> How are they related? When is the solution impossible? CONS TR UCTIONS. 129 When the angles are given, apply the construction of Problem VII. How many solutions are possible ? What kind of A ? B. When three consecutive parts (two sides and included angle or two angles and included side) are given. Solution. Apply the construction in Problem VII. C. When opposite parts (two angles and an opposite side or two sides and an opposite angle) are given. Solution. Apply the construction in Problem VII. When is the construction ambiguous? D. When two sides and the altitude to the third side are given. Solution. Through one end of the altitude draw a normal to it for the base ; about the other end C as centre, with the sides as radii, draw circles cutting the base at A and A\ B and B' ; then ACB or A'CB' is the A required. Why? E. When two sides and the medial of the third side are given. A m Fig. 104. 130 GEOMETRY, \i SA be the medial of BC, and SA^ be symmetric with (Fig. 104) SA as to S, then ABA'C is a parallelogram (why?) ; hence Solution. Take a tract the double of the medial. About its ends as centres with the sides as radii draw circles and then complete the construction. How many A fulfilling the conditions are possible ? How are they related ? F. When the three medials are given (Fig. 105). Solution. Remember that the medials trisect each other ; construct the A OBC according to (E), and draw OA counter to OM and twice as long. 151. Problfem X. — To construct an angle of given size and subtended by a given tract. Data : O the given angle, AB the given tract (Fig. 106). Solution. Construct the angle BAD of given size (how?), draw the mid-normal of AB, meeting AD at P; also the normal to AD at A, meeting the mid-normal at S. About CONS TR UC TIONS. 131 .S" as a centre with radius SA draw a circle ; it touches AD at A (why?). The vertex V of the required angle may be anywhere on the arc A VB or on its symmetric A V^B (why ?) . Fig. io6. 152. Problem XI. ray. To draw a circle tangent to a given Solution. About any point S with a radius equal to the distance of ^S from the ray, Z, draw a circle ; it will be a circle required (why ?) . If the circle must touch the ray L at a given point P, then S must be taken on the normal to L through P (why ?) . If, besides, the circle must go through a given point Q, then ^ must also be on the mid-normal of PQ (why ?) . Hence the construction. 132 GEOMETRY. 153. Problem XII. — To draw a circle touching two given rays. The centre may be anywhere on either mid-ray (why?). If now the circle is to touch a third given ray, the centre must be also on another mid-ray ; that is, it must be the intersection of two mid-rays of the three angles of the three rays. There are four such intersections — what are they? Complete the construction. See Fig. 60. 154. Problem XIII. — To draiv a circle through two points. The centre .S may be anywhere on the mid-normal of the tract AB between the points (why?), the radius is — what? If now the circle is to pass through a third point C, then S must also be on the mid-normal oi BC and CA (why?). There is one, and only one, such point (why?) ; complete the construction. When is the construction impossible ? 155. Problem XIV. — To draw a circle through two given points and tangent to a given ray ; or, tangent to two given rays and through a given point. This double problem is mentioned here because it must naturally present itself to the mind of the student ; but the solution involves deeper relations than we have yet explored. See Art. 000. Several of the foregoing problems were indefinite, admit- ting any number of solutions : these latter taken all together form a system or family. Problems concerning parallelo- grams and other 4-sides may often be solved on cutting the 4-side into two A. 156. Problem XV. — To inscribe a regular 4-side (square) in a circle (Fig. 107). CONS TR UC TIONS. 133 Solution. Join consecutively the ends of two conjugate diameters. The 4-side formed is inscribed (why?) and is a square (why ?) . Fig. 107. 157. Problem XVI. — To inscribe a regular 6-side in a circle. Solution. Apply the radius six times consecutively as a chord to the circle (Fig. 108). The figure formed will be the regular 6-side (why?). Fig. 108. N.B. This seems to have been one of the first geometric problems ever solved. The Babylonians discovered that six radii thus applied would compass the circle, and having 134 GEOMETRY. already divided the circle into 360 steps, they accordingly divided this number by 6 and thus obtained 60 as the basis of the famous sexagesimal notation, which long domi- nated mathematics and still maintains its authority un- diminished in astronomy aiid chronometry. In more difficult problems it is often advisable, or in fact necessary, to suppose the problem solved, the construction made, and investigate the relations thus brought to light. Then the facts thus discovered may be used regressively in making the construction required. This method is illus- trated in the following : ' 158. Problem XVII. — To draw a square with each of its sides through a given point. Let A, B, C, D, be the four given points, and suppose (Fig. 109) FQRS to be the square properly drawn. Draw iO IQ G '^>< S /' F IG, 109. AB, cutting a side of the square, and through B draw BE parallel to the side cut. Through a third point C draw a normal to AB, meeting QR in F. Also draw FG parallel to FQ. Then the A ABE and CFG are congruent (why ?) . CONS TR UC TIONS. 135 Hence we discover that CF— AB. This fact is the key to the Solution. Join two points A and B ; from a third, C, lay off CF equal and normal to AB. The join of D, the fourth point, and 7^ is one side of the square in position (why?). Let the student complete the construction and Show that four squares are possible. 159. Problem XVIII. — To trisect a given angle. Suppose the problem solved and the ray OT to make -^^TOB^^TOA {Y\%.\\o). /w Fig. From any point A of the one end of the angle draw a parallel and a normal to the other end ; also draw to the trisector a tract AS = OA. Then the following relations are evident : -^AOS^-^fASO^^-^ SAT^^ STA) but '^A0S=2'^. TOB = 2 STA ; hence '^STA = ^ SAT, and ST= SA. 136 GEOMETRY. Again, 2C SAR ■= ^ SRA, being complements of equal angles ; hence SA — SR, TR = 2 OA. Hence Solution. From any point A of either arm of the given angle draw a parallel and a normal to the other arm ; then, with one point of a straight-edge fixed at the vertex O, turn the edge until the intercept between the normal and the parallel equals 2 OA. But to do this we need a graduated edge, or a sUding length 2 OA on the edge itself. Accord- ingly, this construction, while simple, useful, and interesting, is not elementary geometric in the sense already defined. To discover such a solution for this famous problem, has up to this time baffled the utmost efforts of mathematicians. EXERCISES II. 1. State and prove the reciprocals of Exercises 9, 10, II, page 71. 2. Find a point on a given ray, the sum of whose dis- tances from two fixed points is a minimum. 3. The same as the foregoing, difference supplacing sum. 4. A and A\ B and B\ C and C\ are symmetric as to MN. Show that AABC=A A^B'C. 5. The inner and outer mid-rays of the basal angles of a symmetric A form a kite. 6. The inner mid-rays of the angles of a trapezium form a kite with two right angles. 7. The joins, of the mid-points of the parallel sides of an anti-parallelogram, with the opposite vertices, form a kite. 8. The mid-rays of the angles at the ends of the trans- verse axis of a kite cut the sides in the vertices of an anti- parallelogram. EXERCISES II. 137 9. How must a billiard ball be struck so as to rebound from the four sides of a table and return through its original place ? 10. Trace a ray of light from a focus F, to another given point Qf reflected from a convex polygonal mirror. 11. A ray of light falls on a mirror M, is reflected along ^ to a second mirror M\ is thence reflected along T. Re- membering that the angle of incidence equals the angle of reflection, show that the angle between the original ray R and its last reflection T is twice the angle between the mirrors (angle RT= 2 angle MM^). On this theorem is grounded the use of the sextant. 12. Two mirrors stand on a plane and form an inner angle of 60° ; a luminous point /'is on the mid-ray of this angle (or anywhere within it) ; how many images of /'are formed? How are they placed? What if the angle of the mirrors be i/n of a round angle? This theorem is beautifully illustrated in the kaleidoscope. 13. A regular /z-side has n axes of symmetry concurring in the centre of the ;/-side, which centre is equidistant from the sides of the //-side. 14. How do these axes lie when 71 is even? when n is odd? Show that if n be even, the centre is a centre of symmetry. 1 5 . The half-rays from centre to vertices of a regular «-side form a regular pencil of n half-rays, and those from the cen- tre normal to the sides, another regular pencil ; also the half- rays of each pencil bisect the angles of the other. 16. In a figure with two rectangular axes of symmetry each point, with three others, determines a rectangle, and each ray, with three others, a rhombus. 138 GEOMETRY, 17. Find the axes of symmetry of two given tracts. 18. A regular A, along with the figure symmetric with it as to its centre, determines a regular 6-angle (6-pointed star) . 19. Two congruent squares, the diagonals of one lying on the mid-parallels of the other, form a regular 8-angle ; also find the lengths of the intercepts at the corners. 20. The outer angle of a regular ?z-side is m times the outer angle of a regular w/z-side. EXERCISES III. 1. A circle with its centre on the mid-ray of an angle makes equal intercepts on its arms. 2. Tangents parallel to a chord bisect the subtended arcs, and conversely. 3. Tangents at the end of a diameter are parallel. 4. A and B are ends of a diameter, C and D any other two points of a circle ; E is on the diameter, and angle AED = 2 angle CAD ; find the possible positions of E. 5. From n points are drawn 2 n equal tangent-lengths to a circle ; where do the points lie ? 6. In a circumscribed hexagon, or any circumscribed 2 «-side, the sums of the alternate sides are equal. 7. If the vertices of a circumscribed quadrangle, hexagon, or any 2 ;?-side, be joined with the centre of the circle, the sums of the alternate central angles will be equal. 8. The sums of the alternate angles of an encyclic 2n- side are equal, namely, each sum is {n — \) straight angles. 9. The joins of the ends of two parallel chords are sym- metric as to the conjugate diameter of the chords. EXERCISES III. 139 10. A centre ray is cut by two parallel tangents. Show that the intercepts between tangent and circle are equal. 11. Normals to a chord from the ends of a diameter make, with the circle, equal intercepts on the chord. 12. The joins of the ends of two diameters are parallel in pairs, and form a rectangle, and meet any two parallel tan- gents in points symmetric in pairs as to the centre. 13. The joins of the ends of two parallel chords meet the tangents normal to the chords in points whose other joins are parallel to the chords. 14. A chord AB is prolonged to C, making BC — radius, and the centre ray CD is drawn ; show that one intercepted arc is thrice the other. 15. The intercepts, on a secant, of two concentric circles are equal. 16. A chord through the point of touch of two tangent circles subtends equal central angles in the circles. 1 7. Two rays through the point of touch of two tangent circles intercept arcs in the circles whose chords are parallel. 18. The transverse joins of the ends of parallel diameters in two tangent circles go through the point of tangence. 19. Four circles touch each other outerly in pairs : ist and 2d, 2d and 3d, 3d and 4th, 4th and ist ; show that the points of touch are encyclic. 20. Show that three circles drawn on three diameters OA, OB, OC intersect on the sides of the A ABC. 21. Find the shortest and the longest chord through a point within a circle. 22. In a convex 4-side the sum of the diagonals is greater than the sum of two opposite sides, less than the sum of all the sides, and greater than half the sum of the sides. 140 GEOMETRY. 23. Three half-rays trisect the round angle O \ on each is taken any point, as A, B, C. Find a point M such that the sum AfA + MB + MC is the least possible (a minimum). 24. Two tangents to a circle meet at a point distant twice the radius, from the centre ; what angles do they form ? 25. The intercept of two circles on a ray through one of their common points subtends a constant angle at the other. 26. What is the envelope of equal chords of a circle? 27. Two movable tangents to a circle intersect under con- stant angles ; find the envelope of the mid-rays of these angles. 28. The vertex Fof a revolving right angle is fixed mid- way between two parallels, and its arms cut the parallels at A and B ; find the envelope of AB. 29. From a fixed point B a normal BN is drawn to a movable tangent 7" of a circle, and through the mid-point M of /W there is drawn a parallel to T; find its envelope. 30. The vertices of a A are Fj, Fg, Fg ; the mid-points of its sides are M^, M,,, J/i ; the feet of its altitudes are Ai, A2, As ; the inner bisectors of its angles meet the oppo- site sides at B^, B.,, B^; and the outer bisectors at B\, B'2, B\ ; its centroid is C, its in-centre is /, its circum- centre is S ; its angles are ai, a^, a^, and their complements are a\, a'2, a\. Express through these six angles the angles between : (i) Fi^iandFiF; (2) A^A^Sind Fz^s', (3) AiAoamd V-^A^; (4) A^Az a.nd A^A^ ; (5) MiA^ a.nd V^V,; (6) J/i^sand K.V^; (7) J/j^s and J/^^a ; {d>) A.M^ and A^M^; (9) A^M., and A^Ao^; (10) SV^ and SV^] (11) SV^ and V,V,', (12) SV^ and V^A^; (13) IV^ and JV\', (14) /Fiand V^A.,; (15) F^'j and V^B'^. EXERCISES IV. 141 31. Find the locus of the mid-points of chords through a fixed point upon, within, or without a fixed circle. 32. Find the locus of the mid-points of the intercepts of a secant between a fixed point and a fixed circle. 33. As the ends of a ruler slide along two grooves normal to each other, how does its mid-point move ? 34. Two equal hoops move along grooves normal to each other and touch each other ; how does the point of touch move? 35. Orthocentre O, centroid C, circum-centre 6", and cen- tre F of Feuerbach's (9-point) circle, of a A are collinear, and 0C=2CS (Euler), OF=FS. 36. Two parallel tangents meet two diameters of a circle at the vertices of a parallelogram concentric with the circle. 37. The inner mid-rays of the angles of a 4-side form an encyclic 4-side. 38. The outer mid-rays of the angles of a 4-side form an encyclic 4-side. How are the 4-sides of 37 and 2>^ related? 39. The circum-centres of the four A into which a 4-side is cut by its diagonals are the vertices of a parallelogram. 40. The circum-centres of the two pairs of A, into which a 4-side is cut by its diagonals in turn, are how related to each other and to the centres in 39 ? EXERCISES IV. 1. Construct a square, knowing {a) Its side ; or (6) its diagonal. 2. Construct a rectangle, knowing {a) Two sides ; or {d) a side and a diagonal ; or {c) either a side or a diagonal and the angle of either with the other ; or {d) a diagonal and its angles with the other diagonal. 142 GEOMETRY. 3. Construct a parallelogram, knowing {a) Two sides and one angle ; or {b^ a side, a diagonal, and the included angle ; or (r) two sides and the opposite diagonal ; or (^) two sides and the included diagonal ; or (EH or BC ; they cut the whole square into four parts, namely, the square on a, the square on l>, and the two congruent rectangles ad and ab. q. e. d. Corollary. The square on a tract equals four times the square on half the tract. 158 GEOMETRY. [Th. XCIII. i8i. Theorem XCIII. — The square on the difference of two tracts equals the sum of the squares on the tracts, less twice the rectangle of the tracts. Proof. In the preceding figure treat AB ox a -\- b z.% the one tract and EB or b as the other \ then AE or a is the difference. The square ^ C is the square on the tract a -{- b \ increase it by the square PC on b, so that this square is to be counted twice and thought as doubly laid in the figure ; now strip off the rectangle HC or (^ + ^) b, and its congruent BG\ there remains the square AP on the difference a. Q. E. D. 182. Theorem XCIV. — The rectangle of the sum and dif- ference of two tracts equals the difference of the squares on those tracts. Proof. In the same figure treat (^ + <^ as the one tract and a as the other, so that 2 ^ + /^ is the sum and b the dif- ference of the tracts. Then the two rectangles ^Cand HG agree in one dimension b and the sum of their other dimen- sions is 2 ^ -f- /; ; hence their sum is the rectangle {2 a -{- b')b ; i.e. the rectangle of the sum and difference of the tracts a-\-b and a. Moreover, this area is plainly what is left on taking away the square on a from the square on {a-^ b) -, hence, etc. q.e.d. Scholium. Calling the tracts u and v, we may express these theorems in symbols thus : (// -f z')- =// -f ^^" + 2 uv ; {u — vY = u^ -{-7>^ — 2 uv ; {u 4- 1!) {u — v) = u^ — vK But let the student carefully beware of importing any algebraic meaning into these symbols at this stage of the discussion ; u'^, for example, does not mean the product of u multiplied by u, but merely the square whose side is //. Tk. XCV.] SQUARES. 159 183. Theorem XCV. — The square on the hypotenuse of a right A equals the sum of the squares on its other sides (Fig. 125). h C h C i ^ / i / 1 ,--"'" \ \ A' Fig. 125. Proof. Let AC and A^C be two congruent squares on the tract a -\- b -, take away from each the four congruent right A, I, 2, 3, 4 ; there is left of the one the square on the hypotenuse of the right A, and of the other the sum of the squares on the legs, a and b. Q. e. d. Scholium. This most famous theorem was discovered by Pythagoras (circa B.C. 550), it is said, while pursuing cer- tain arithmetical researches. He was, in fact, seeking out pairs of numbers, the sum of whose squares was itself the square of a number, when he made the astonishing observa- tion that all such pairs, when measured off in terms of a unit length, formed the two legs of a right A of which the third number, similarly laid off, formed the hypotenuse. Proofs of the proposition abound. In the classic one of Euclid, it is shown that the square on ^C= rectangle AD (Fig. 126), the media of comparison being the congruent A BAE and FAC. Similarly the square on ^C= rectangle -5Z>. The 160 GEOMETRY. [Th. XCV. demonstration in the text seems to be the most simple and direct that is possible, while its presuppositions are the least possible. Fig. 126. 184. Naturally we now inquire into the relations among the squares on the sides of oblique A. Def. The foot of the normal, from a point to a ray, is called the (orthogonal) projection of the point on the ray ; and the tract between the projections of the ends of a tract is called the projection of the tract itself (Fig. 127). Thus /" and Q are the projections of /'and Q on Z, and P^Q is the projection oi PQ on L. Th. XCVL] SQUARES. 161 185. Theorem XCVI. — The square on any side of a tri- angle equals the sum of the squares on the other sides, in- creased or decreased by tivice the rectangle of either and the projection of the other upon it, according as the first side lies opposite an obtuse or an acute angle. Data: ABC the A, BD the projection of BC on AB (Fig. 128). C Proof. and hence But hence AC^AEC^ CD\ CD' =BC-- BD- (why ?) ; A C- = AD- - BD'- + BC\ AD- - BD = AB {AB + 2 BD) (why ?) ; ^ C^ = AB^ +BC'^2 AB'BD. Thus far ji has been considered obtuse ; if it be acute, then AD= AB + BD ; but BD is to be reckoned leftward, oppositely to BD in the former case ; that is, BD is reversed in sense, a fact which we may express in symbols by writing AD = AB- DB. Hence results, as before, AC^ = AB- + BC- ~2AB' DB. q.e.d. 162 GEOMETRY. [Th. XCVII. Scholium. We observe that when /8 is a right angle, then, and then only, D falls on B, the projection on BD vanishes, and there results the Pythagorean Theorem, AC^AB'^^BCK As /3 changes from obtuse to acute, the point B passes from the left to the right of D, and the tract BD changes its sense, — from being reckoned rightv^dsdi it comes to be reckoned left^'dxA. It is this change of sense in BD that changes the addition into the subtraction of the rectangle. It is often absolutely necessary to take account of the sense of a magnitude, the way it is reckoned, in order to perceive the generality, the internal coherence and continuity, of our results. Corollary. When the square on one side of a A equals the sum of the squares on the others, the A is right-angled opposite that side. Converse of the Pythagorean Theorem. 1 86. Theorem XCVII. The sum of the squares on two sides of a A equals twice the sum of the squares oti half the third side and its medial. Data: ABC the A, CM medial, and CN normal to AB (Fig. 129). Proof. AC'' = am'' + CaP -f 2 AM- MN, BC' = ¥M^ + CM''-2BM'MN. Th. XCVIII.] SQUARES. 163 Adding and remembering that AM= BM, we get AC \BC =2{AM -{- CM). Q. E. D. Corollary i. If a, b, c be the sides of the A, and tn the medial of c, then 4;;r 2a^ 2 b' Corollary 2. If the A be regular, then ;// is the altitude, and if J- be a side, *i87. Theorem XCVIII. — The sum of the squares on the sides of a quadrangle equals the sum of the squares on the diagonals and four times the square on the join of the mid- points of the diagonals. Data: ABCD the 4-side, ^i^ the join of the mid-point of the diagonals (Fig. 130). C Proof. whence Fig. 130. AR ^Alr =2AJ^'^ -^2 ^C' + C^'=2^^'+2C^' (why?); AB^ 4- BC' + ~CIX + Djt = ^BF^-{-2AF'' + 2 W = ^BF^ + /[AE- + ^EF' (why?) ^mf^-AC^'-^AEF" (why?). q.e.d. 164 GEOMETRY. [Th. XCIX. Corollary. The sum of the squares on the sides of a O equals the sum of the squares on the diagonals. (Why?) 1 88. Theorem XCIX. — The diffej-ence between the squares on a side of a symtnetric A and on the join of the vertex to any point of the base equals the rectangle of the segments into which that point divides the base. Data : ABC the symmetric A, F any point of the base (Fig. 130- P A . P' Fig. 131. Proof. Let J/ be the mid-point of the base, then BT = BA' 4- AB' +2AP' AM (why ?) . .-. bP -BA'' = AF\AF+2AM\ = AF' PC. q.e.d. Now let F move towards ^4 ; as it reaches A, the tract AF vanishes, and so do both sides of the equation. As F moves on to F^ towards C, the tract AF changes sense, it is no longer reckoned leftward, but rightward ; at the same time the left-hand side of the equation chaiiges sign, BF becoming less than BA ; but the equation still holds, for the sign of AF must also change with the change of sense. We are yet at liberty to choose the difference of squares as either ^F" -K4^ or ^A^ -'BF\ The first is perhaps preferable, and we see that when F is without the tract A C, then FA and FC have the same sense and sign, being reckoned the same way, and the difference is positive ; but Th. C] squares. 165 when P is within A C, then PA and PB have opposite sense and sign, being reckoned oppositely ; hence we may say their rectangle is negative, and accordingly BP is less than BA. It is extremely important to note that an area is a sign- magnitude^ positive or negative ; it has sense. 189. When P is at A, the difference BP' — Bj^ is o ; as P moves towards C, the tract BA remains unchanged, but BP shortens until P reaches M \ thence BP lengthens until it again becomes equal to BA or BC, as P falls on C. Hence the difference BA — BP', or its equal PA • PC increases while P moves from A \.o M and decreases as P moves from M to C \ hence it is greatest when P is at M. A value of a variable magnitude that is thus the greatest within a series of successive values, or that is greater than the values just before it and just after it, is called a maxi- mum ; while a value that is less than the values next before and next after it, is called a minimum. Hence the rectangle AP' PC is a maximum for Psit M; or the rectangle on the two parts into which a given tract may be divided is a maxi- mum when the parts are equal, or 0/ all rectangles with a given perimeter, the square is the maximinn. Once more, AC =AP-\- PC = AP- + PC + 2 AP-PC. Now ^ C is constant while P moves from A to C, and AP- PC is greatest when P is at M ; hence AP' + PC^ is least when P is at M ) that is, the sum of the squares on the two parts of a given tract is a ?ninimum when the parts are equal 190. Theorem C. — The rectangle of the distances on a secant from a fixed point to a fixed circle is constant for all secants. Data: P\\\q. fixed point, 6* the fixed circle (Fig. 132). 166 GEOMETRY. [Th. C. Proof. Through /* draw any two secants cutting the circle at A and B and at C and D. Then in the symmetric AAOB, OP-- '0A'=PA ' PB and in A COD, ~0P' -~0C- = PC'PD (why ?) . Hence PA • PB = PC - PD (why?), no matter how the secant be drawn through P. q. e. d. Fig. 132. Def. This constant, namely, the area of the rectangle of the distances from the fixed point to the fixed circle along any secant, is called the power of the point as to the circle. Corollary i. For a point within the circle the power is the square on half the shortest chord through the point, or on half the chord through the point normal to the radius through the point (why ?) ; for a point without the circle the power is the square on the tangent-length from the circle to the point (why?) ; for a point on the circle the power is zero (why ?) . Corollary 2. The power of a point without the circle is positive ; of a point within, it is negative (why?). Corollary 3. If PT' — power of P as to S, and T be on S, then PT\^ tangent to S (why?). Th. CL] POWER-AXIS. 167 Corollary 4. If in be the minimum distance from the point F to the circle 6* of diameter d, then the power of the point is m(d ± m) according as F is without or within S. This notion of the power of a point as to a circle is so ex- ceedingly important that it may be well to exemplify its use, in passing, though not necessary for our present purposes. 191. Theorem CI. — All points having equal powers as to two circles lie on a ray. Data: 6" and S the two circles (Fig. 133). Fig. 133. Proof. Draw a normal to the centre-tract 00^ (atiV), cutting it into two parts d and d\ and from any point F on this normal draw tangents FT, FTK Then r and / being the radii, FN'--^d^=FT^+r', and FN'^-^t d'''=F¥'^+ r^"" (why?). Hence d'- - d'- = FT""- FT''^-^ r^ - r'\ Hence FT = FT'^ when and only wheii d^—d^^ = r^—r'^. If then we find iV^on 00', dividing it so that d"^ — d'- = ^2 _ ^'2^ ^jj^ ^i^jg ^,^j^ always be done, then the powers of all 168 GEOMETRY. [Th. CI. points on the normal through N will be equal, and the powers of all points not on this normal will be unequal. Q. E. D. 192. Def. This most important ray was discovered in 18 13 by Gaultier and named by him i-adical axis of the two circles ; a better name would seem to be power-axis. This discovery marked and in a measure determined the renas- cence of Geometry. Corollary i. The common secant of two circles is their power-axis. Corollary 2. The common tangent of two circles is their power-axis. Corollary 3. The power-axes of three circles taken in pairs concur. Def. The point of concurrence is named radical centre or power-centre of the three circles. Corollary 4. A circle about the power-centre with the common tangent-length as radius intersects the three circles orthogonally. The importance of the following discussions can scarcely be overestimated. They are meant to ground firmly and in strict geometric fashion the doctrine of PROPORTION. 193. If P be any point not on a circle S, FT a tangent, and FAB a secant of S, then we have seen that FA' FB = FT^ for all directions of FAB. If a circle / be drawn about F with radius FT, it will cut ^ orthogonally (why ?) ; then A and B are called inverse points as to F, which is called the centre of inversion, while / is called the circle of inversion. Th. CIL] proportion. 169 194. Theorem CII (converse of CI). — If the rectangle of distances from a point to two points on a ray equal {in sense as well as sign) the rectangle of the distances from the point to tivo poifits on another ray, both rays going through the point, then the tivo pairs of points are encyclic. Data : P the point, A and B, C and D, the two pairs of points, and PA • PB = PC - PD (Fig. 134). Proof. The circle through A, B, and C will meet PD somewhere, as at D\ Then PA - PB = PC - PD' (why ?) ; hence PD = PD' (why?), or D' is D. Query. Where and why is it necessary to regard the sense of the rectangles in this proof ? 195. Now let us consider this encyclic quadrangle. We know that the opposite inner ^s, as A and D, are supple- mental (Fig. 135). Think of the plane as a doubly laid film with AC drawn in the lower and BD drawn in the upper layer, and imagine PBD taken up, turned over, and 170 GEOMETRY. [th. cm. replaced so that B will fall on B^ and D on D' . Then will B^D^ be II to AC. For the inner angle at Z>' equals inner angle at D ; hence the inner ^s at A and Z>' are supple- mental (why?) ; hence B^D^ and AC sue II. Fig. 135. This operation of turning over BD into the position B'D' we may call inverting BD. Hence Theorem CIII. — If one side of an encyclic quadrangle be inverted^ the resulting figure is a trapezoid. Conversely, 196. Theorem CIV. — If 07ie parallel side of a trapezoid be inverted J the figure resulting is an encyclic quadrangle. The ready proof is left for the student. We note that the proofs hold as well for the crossed quad- rangle and trapezoid as for the convex or normal. 197. Now let PAC and PB^D^ be any two A with com- mon vertical angle P and II bases A C and B^D ; then PA . PB' = PC . PD' (Fig. 135). Th. civ.] proportion. 171 Proof. For on inverting B^ D^ into BD the quadrangle ABCD is encyclic. Hence PA • FB = PC - PD, and PB = PB', PD = PD\ Hence q. e. d. 198. Conversely, Let PAC and PB^D^ be any two A with common vertical ^ and let PA ' PB' = PC • PD\ Then AC and B'D' are II. Proof. Draw through A a II ^^C to B'D\ cutting PB' at C; then PA . PB' = /^r • PD\ Hence /'C:= /'C (why?). q.e.d. 199. These relations are very simple and easy of com- prehension, but their statement in words is very awkward and cumbrous. To relieve the difficulty of verbal expression we introduce a new arbitrary definition and a new arbitrary symbolism. Def. If the rectangle of two tracts equals the rectangle of two other tracts, the four tracts are said to be in propor- tion, or to form a proportion, or to be proportional. Symbolism. If ;/ and v be the one pair, x and J^' the other pair of tracts, then we write u : x: :y : v and read u is to X as y is to v. 200. In order to speak readily about this proportion we define further : Definition i. The tracts are called terms of the propor- tion. Definition 2. The first and last are called extremes; the second and third are called means. 172 ^ GEOMETRY. [Th. CIV. Definition 3. The first and second are called the first couplet ; the third and fourth, the second couplet. Defifiition 4. The first and third terms are called antece- dents; the second and fourth are called consequents. Definition 5. When the two means are equal, each is called the mean proportional or geometric mean of the extremes. Definition 6. The fourth term is called the fourth propor- tional to the other three taken in order ; or, if the means be equal, it is called a third proportional to the other two taken in order. Definition 7. When the means are exchanged, or the extremes are exchanged, the proportion is said to be alter* nated. Definition 8. When the terms of each couplet are ex' changed, the proportion is said to be inverted. Definition 9. When in place of the first or second term of each couplet is put the sum (or difference) of the terms of that couplet, the proportion is said to be compounded (or divided) . 201. Since by a proportion we mean nothing more and nothing less than that the rectangle of the means equals the rectangle of the extremes, it is plain that the same proportion may be written in several different ways, thus : u \ X \ \v '. y, u \ v\ '. X : y, y\ X \ w \ u, y \v \ \ X \ u, X \u \\y w, X'.ywu'.v, v \ u : \y \ x, v: y \ \ u : x, all mean precisely the same ; namely, rectangle of // and y = rectangle of v and x. 202. All of these forms, and no others, may be derived from any one of them by alternation and inversion ; hence Th. CVL] PROPORTION. 173 Theorem CV. — When four tracts are in proportion they are in proportion by alternation and by inversion {alternando et invej'tendo). The simpUfication of expression will now soon become evident. We must still further premise, however, Definition lo. When the angles of one A are respectively equal to those of another, the A are said to be mutually- equiangular, and the sides opposite equal angles are said to correspond, as do also the equal angles themselves. 203. Theorem CVI. — Corresponding sides in two mutu- ally equiangular A aj'e proportional in pairs. Data : ABC, A^B'C the A, A = A', B = B', C= C\ Proof. Fit ^ ^ on ^ ^' ; then ^C is II to B' O (why?). Hence AB 'A'C' = A'B' • AC (why?), or AB: A'B':: AC: A' a. C Fig. 136. Similarly, by putting B on B', C on C, BC:B'C ::BA:B'A', CA : a A' ::CB: CB', Q. E. D. 204. These three proportions may be conveniently written as a continued proportion, thus : 174 GEOMETRY. [Th. evil. AB'.BC'.CA:: A'B' :B'C': C'A', read AB is to BC is to CA as A'B' is to B'C is to C'A' ; or perhaps still better thus : AB : A'B' ::BC:B'C'::CA: C'A', read AB is to A'B' as BC is to B'C as C^ is to C'A' ; they are exactly equivalent to the three equations AB.B'C' = A'B' 'BC, BC- C'A' = B'C- CA, CA-A'B'= C'A' -AB. 205. Theorem CVII. — Conversely, Two Awt'fk sides pro- portional are fnutually equiangular. Data: ^^C, ^'.5' C the two A, and AB.A'B'.-.BC'.B'C'.'. CA-. C'A' (Fig. 137). C C Proof. From C lay off CA equal to CA' and draw A^B^ II to AB. Then CA^ ' CB=CA' CB^ (why?). But CA' ' CB=CA' CB' (why?). Hence CA - CB^ = CA - CB' (why?). Hence CB, = CB'. Similarly, A^B, = A'B'. Th. CIX.] PROPORTION. 175 Hence A^B^C and y^j^iCare congruent (why?). Hence A^B^ C and AB C are mutually equiangular (why ?) . Q. E. D. Thus it appears that mutual equiangularity and propor- tionality of sides coexist and imply each other. Two such A mutually equal in their angles and proportional in their sides are called similar. 206. Theorem CVIII. — T^vo A having an^of one equal to an ^ of the other and the including sides proportional are similar. Data : ABC, A'B'C the two A, ^C= ^a, and CA : C'A' : : CB : C'B' (Fig. 138). Proof. Fit C on C; then by the proportion A'B' is II to AB. Hence, etc. q. e. d. 207. Theorem CIX. — Tiifo A having two pairs of sides proportional, and a pair of angles opposite the larger sides in each equal, are similar. Data : ABC, AB^C the two A, AB>BC,A'B^>B'a. AB\A'B' wBC'.B'C, and ^C=^C' (Fig. 139). 176 GEOMETRY. [Th. ex. Proof. Fit ^ C on C ; then since A'B' > B^C and AB > BC, both ^^ and AB' must be drawn making the inner ^s at A and A^ acute. From the proportion, AB and ^'Z?' are now li (prove it) ; hence the A are mutually equiangular and hence similar. Q. E. D. N.B. If AB were < BC and A'B' < B'C, then AB and A'B' might make the ^s at A and ^' supplemental instead of equal, and hence AB and A^B^ anti-ll instead of II, in which case the A would not be similar. Dra^ a figure illustrating this case. Compare the conditions of similarity with the conditions of congruence between two A. 208. Theorem CX. — If two proportions agree in the first three terms of each, they agree ifi the fourth also. Data : u :x: -.7! : y, and u\ x\\v\ y'. Proof, uy = vx, and uy' = vx : hence uy = uy\ Hence j' = jv' (why?), q. e. d. 209. Theorem CXI. — If two proportions agree in one couplet, the other couplets form a proportion. Th. CXIL] proportion. 177 Data : u\x'.'.v\y\ u-.xww.z (Fig. 140) . Proof. Suppose w < v and on two half-rays through P lay oKFU, FX, FV, and FY:= u, x, v,y. Open the angle at F until UV= w. This is possible, since w be the inner centre, they may be said to be in indirect perspective, or in contra-perspective. If on any ray through O there be taken two points, C and C\ such that 0C\ OC \\t\t\ then C and C are said to correspond to each other with respect to the centre of simili- tude O in the ratio of similitude / : /' ; any tract between two points in the one figure is said to correspond to the tract between the corresponding points in the other figure. 227. Theorem CXXIII. — Coi-respondent tracts in two perspective figures are parallel and in the ratio of similitude to each other. Data : O the centre, A and A\ B and B^ two pairs of corresponding points (Fig. 151). Th. cxxv.j SIMILAR FIGURES. 189 Proof. The A A OB and A^OB' are similar (why?); hence AB and A'B^ are II, and AB : A^B' ::t:t' (why?). Q. E. D. P' Fig. 151. 228. Theorem CXXIV. — Conversely, If from two points, A and A', correspondent as to O, there be laid off two II tracts AB, AB^ in the ratio OA : 0A\ then B and B' cor- respond. Let the student give the proof. 229. Theorem CXXV. — Any two circles are in perspec- tive and contra-perspective. Data: 6" and S any two circles (Fig. 152). Proof. Divide the centre tract CO innerly and outerly in the ratio of the radii r : r' at points / and O. Draw any secant OA, and on it lay off OA' so that OC: OC:: OA :0A\ The A OCA and OCA' are similar (why?). Hence C'A' = r' (why?) ; hence A' is on S' ; hence any point of 190 GEOMETRY. [Th. CXXVI. .S has its correspondent on 6"' in the same fixed ratio of the radii ; hence 6* and S are similar, and are plainly in perspec- tive. For / the reasoning is the same, but the tracts being laid off oppositely, the figures are in contra-perspective. Q. E.D. Coi'ollary. Common tangents to the two circles go each through a centre of simiHtude. Fig. 152. 230. Theorem CXXVI. — Conversely, Atiy figui-e simi- lar to a cii'cle is itself a circle. Data : S a circle, S similar to it with respect to the cen- tre of similitude O, in the ratio r : r'. Proof. Find the corresponding point C of the centre C of S, and draw through O any secant meeting S and S in the corresponding points A and A\ Then triangles COA and COA' are similar (why?) ; hence OC: OC:: CA : C'A', Th. cxxviii.] similar figures. 191 and since (9C, OC , and CA are constant in length, so, too, is CA\ Hence S is a circle about C as centre. The like proof holds for the inner centre /. q. e. d. 231. Theorem CXXVII. — The angle between two tracts in one figure equals the angle between the corresponding tracts in any similar figure. Data: i^and F^ (Fig. 151), two similar figures similarly placed. AB and BC, two tracts in F. A'B' and B'C, the corresponding tracts in F'. Proof. Draw OA, OB, OA', OB' ; then the theorem fol- lows at once from similar A. But if the figures be not similarly placed, and F" be one of them congruent with F', suppose F" brought to coincidence with F' ; then what has just been proved for F' holds for F". q.e.d. Corollary. F" may be brought to coincide with F' by being mtxtXy pushed, — all rays remaining parallel to them- selves in their original position, until one point of 7^" falls on the corresponding point F\ — and then being merely turned until another point of /^" falls on its correspondent in F'. If the figures still do not coincide throughout, but only on the common ray through three points, it will be necessary and sufficient to revolve F' about that common ray through a straight angle, which revolution will change opposition into superposition of the figures. In this revolution all rays in the figure are turned through the same angle ; hence 232. Theorem CXXVIII. — /;/ two similar figures all lines are inclined to their corresponde7its under the same angle. Perhaps we might name this angle the anomaly of the one figure as to the other. In two similar figures point corresponds to point, angle to 192 GEOMETRY. [Th. CXXIX. equal angle, tract to tr^ct, in the same ratio ; hence it is plain that 233. Theorem CXXIX. — Any two similar figures 7nay be cut up into pairs of similar figures in the sa?fie ratio of simili- tude and order of arrangement. INSTRUMENTS. 234. There are four important instruments used in prac- tice in the construction of similar figures : proportional compasses, sector, diagonal scale, and Pantagraph or Eido- graph. Of these the last is the most interesting and il- lustrates in its working very accurately the definition given above of similar figures in contra-perspective. Every well appointed academy should be furnished with these instru- ments, which may easily be explained and operated. CONSTRUCTIONS. 235. The doctrine of proportion is extensively employed, not only in mechanical drawing with the instruments men- tioned, but also in the strict logical solution of problems of construction. 236. Problem I. — To divide a tract {innerly and outerly) in a give ft ratio, as of I: m. (See p. 182.) 237. Problem II. — To divide a tract {innerly and outerly) into any number of parts proportional to /, ;//, n, p ' • •. Solution. From the beginning of the tract AB draw any half-ray, as AR; on it lay off in order consecutively the tracts /, m, n, p,- • • . Join the end of the last with the end of AB, and through the ends of the others draw parallels ; to this join RB ; they divide AB as required (why?). Let the student solve the problem of outer division. CONS TR UC TIONS. 193 238. Problem III. fwo tracts. To construct the geometric fnean of Solution. On the sum of the two tracts, /and m (Fig. 153), as diameter, draw a circle, and through their common point draw a half-chord conjugate to the diameter ; it is the mean proportional required (why?). Fig. 153. 239. Problem IV. — To construct a square equal to a given rectangle. Proceed as in Problem III. 240. Problem V. — Knowing one dimension of a rectangle equal to a given rectangle, to find the other; or, given three tracts, to find a fourth proportional to them in order. f IG. 154. Solution. Prolong one side of the given rectangle by the given side of the other, as AB to Qy and draw QC meeting 194 GEOMETRY. AD at R; then DR is the other dimension sought (why?) (Fig. 154). Or, On any two half-rays meeting at A lay off AB and AD equal to the given sides or the second and third of the three tracts, and on either, as AD, lay off AQ equal to the one given dimension or the first tract. Draw BQ and DR II to BQ; then AR is the fourth proportional sought (why?). These constructions require us either to know the angle at A or else to draw a parallel. But we may proceed thus, avoid- ing all use of pai^allels and angles : draw a large circle and lay off as a chord of it the difference of the second and third proportionals ; from the ends A and B of this chord lay off AP and BP equal to the second and third proportionals ; about P describe a circle with the first proportional as radius intersecting the circle at /; draw PI, meeting the circle also aty; then ^is the fourth proportional sought (why?) (Fig. 155)- Fig. 155. N.B. The first circle must be drawn sufficiently large, so that the second circle may meet it. 241. Problem VI. — To construct a A equal to a given /Inside. CONS TR UC TIONS. 195 Solution. Drawn either diagonal, as AC, of the 4-side ABCD, and then from D draw a II to the diagonal, cutting AB at A\ Then ACE is the A sought (why?) (Fig. 156). 242. Problem VII. — To construct a A equal to a given n-side. Proceed as in Problem VI, and reduce one by one the number of sides drawn to three. 243. Problem VIII. — To divide a parallelogram into n equal parts by parallels to a side. Solution. Divide an adjacent side into n equal parts and draw parallels ; the n resulting parallelograms are congruent (why?). 244. Problem IX. - — To divide a A into n equal parts by tracts drawn from a vertex. 245. Problem X. — To divide a trapezoid into n equal parts by tracts between and parallel to the parallels. 246. Problem XI. — To divide a A into n equal parts by tracts drawn from a point on a side. Solution. Divide the side containing the point into n equal parts ; from the points of division draw parallels to the join of the point with the opposite vertex ; draw tracts from the 196 GEOMETRY. point to the intersections of these parallels with the other sides ; they are the dividing lines required (why?). 247. Problem XII. — From a point within a A to bisect the A by tracts drawn to a given vertex and to a side (Fig. 157). C M Fig. 157. Solution. If P be the given point, C the given vertex, and PQ the required tract, then on drawing the medial CM it becomes plain that A CPQ = A CMQ ; hence CQ is II to PM. Hence the construction : draw the medial CM, then PM, then CQ II to PM, then PQ. 248. Problem XIII. — Prom a point within a A to bisect the A by two tracts, one of which is drawn to a given point on one side, and the other as may be (Fig. 158). C CONS TR UC TIONS. 197 Solution. If P be the given point within the A, Q the given point on the side, suppose PR to be the required tract. Then on drawing CM and PM and a II to PM through Q cutting CM at /, we have A PQM= A PIM (why?). Also on drawing CP and IR we must have APIC-APRC{^hy}). Hence IR is II to PC (why?). Hence we construct PR (how?). 249. Problem XIV. — To bisect an n-side by a tract drawn from a given vertex. Solution. Let A be the given vertex ; join the adjacent vertices B and Z, and through each of the others draw a tract across the «-side II to BL. Bisect these parallels by a train of tracts from A. This broken line will bisect the «-side (why?), and by Problem VH the student may con- vert it into a single tract from A (how?). 250. Problem XV. — To construct a square equal to the sum of two given squares. Use the Pythagorean Theorem. 251. Problem XVI. — To construct a square equal to the sum of tivo given rectangles. Combine the methods of Problems IV and XV. 252. Problem XVII. — To construct a square equal to 2, 3, 4, • • • « times a given square (Fig. 159). Hint. The diagonal of the given square will be the side of the double square (why ?) ; normal to this diagonal, OB^ lay of[ BC equal to the side of the square ; draw OC, and 198 GEOMETRY. [Th. CXXX^ again normal to it lay off CD equal to the side of the origi- nal square ; draw OD, and so on. In this way we may duplicate, triplicate, ;2-plicate the original square. The broken Hne ABCD • • • and the varying hypotenuse wind round O forever. D 253. Problem XVIII. — To construct a square equal to one half, one third, one fourth, • • • one nth of a give 71 square. Hint. Half the diagonal of the given square is the first side sought ; the altitude of a regular triangle whose side is the given side is the second ; one half of the given side is the third ; • • • in general, the geometric mean between the side and the ^th part of the side of the given square will be the side of the square sought (why ?) . 254. Problem XIX. — To construct a square the Ti\h-fold or the nth. part of a given i^ectangle, parallelogram, or A. Combine the methods of the foregoing problems. 255. Theorem CXXX^. (Lemma). — If a jrc tangle equal a square, and the dimensions of the two be changed propor- tionally {i.e: so that the new and the old dimensions taken in pairs of correspondents form a continued proportion), then the new rectangle will equal the new square. CONSTRUCTIONS. 199 Data : R and R\ two rectangles with dimensions a and b, a} and b\ s and s\ two squares with dimensions t and f ; R = S,2Lnda:a'::b'.b':'.f:f' (Fig. i6o). Fig. i6o. Proof. On the same half-ray from the same point P lay off tracts FA and RB equal to a and b ; on their difference (AB = 2 r) as diameter describe a circle and draw the tan- gent PT: it will equal / (why?). With a radius r' such that a : a' : : r: r' describe a concentric circle, prolong OT to meet this circle at T', and at T' draw a tangent meeting the diametral ray at P. Then PA' = «', F'B' = ^', /^/' = / (why?), and a'b' = /'/ (why ?) . q. e. d. Corollary. If two rectangles (or parallelograms or A) be equal, and their dimensions be changed proportionally, they will remain equal. Prove this corollary in detail, and state it along with the foregoing theorem in symbols. 256. Problem XX. — To construct a rectangle similar to a given rectangle but of double the area. Solution. Construct a square equal to the given rectangle ; then either dimension of the double rectangle will be a 200 GEOMETRY. fourth proportional to the side and diagonal of the square and the corresponding dimension of the given rectangle ; that is, s : d: : a : a^ (why?). Now, however, all squares are similar (why?) ; hence, denoting by d' the diagonal of the square on the side a, we have s : d\ '. a '. d^ ; whence a \ d^ \\ a \ a\ ox a^ ^= d\ Similarly, ^' is the diagonal of the square on the other dimension b. Or we may find h^ by drawing a diagonal of the double rectangle through the end of a^ parallel to the diagonal through the end of a. 257. Problem XXI. — To construct a rectangle similar to a given rectangle but of n-fold the area. Solution. If we construct a square, of side s, equal to the given rectangle and also a square, of side j-', equal to the. re- quired rectangle, then if a and a^ be corresponding dimen- sions in the two rectangles, we have s: aw s^ : a^ (why?). Now, however, if we construct, according to Problem XI, two broken lines, one on j- as a basis, the other on a as basis, the two will be similar figures (why ?) ; so that if j"„ and ^„ be corresponding hypotenuses in the two figures, we have Hence i-' : ^' : : s^^ : a^ ; hence, if s^ = s\ then a^^ = a\ Accordingly, we find «' by constructing (Problem XI) the side of a square the n-{o\A of the square on a. The con- struction is then completed (how?). CONSTRUCTIONS. 201 258. Problem XXII. — Let the student extend the same methods to the co7istruction of parallelograms and 3\ similar to giveti ones but of 71-fold area. 259. Problem XXIII. — To construct a rectangle {paral- lelogram or A) similar to a given one but of one half, one third • • • one n\h the area. 260. Problem XXIV. — To construct a figure similar to a given figure bu^t of double, triple, • • • n-f old area. Solution. On any tract in the figure construct a square, then construct another square, of double, triple, • • • /z-fold area ; its side will be the tract in the new figure corresponding to the assumed tract in the original figure (why?). All other points and lines in the required figure may now be found by drawing parallels. Let the student carry out the construction. 261. Problem XXV. — To construct a figure similar to a given figuj-e but of one half, one third, • • • one nth. the area. The solution is like that of Problem XVIII, mutatis mu- tandis. 262. We have learned to inscribe in a circle a regular 3-side, 6-side, i2-side, • • • 3'2"-side, also a regular 2"-side, and it is natural to inquire how to inscribe a regular 5 -side, lo-side, • • • , 5-2^-side. As it was easiest to begin with the 6-side, so it is easiest to begin with the lo-side ; to inscribe the 5 -side directly presents difficulties. 263. Problem XXVI. — To inscribe a regular 10-side in a circle, suppose the problem solved and AB the side sought. Then in the symmetric A A OB the vertical ^ is half of either basal angle (why?) (Fig. 161). Hence, if we draw AC bisecting angle A the A AOB and ^^C will be similar (why?). 202 GEOMETRY. Hence OB . AB : : AB \ BC, ox OB : OC : : OC : BC. Hence, in order to find AB or OC, it is necessary to divide the radius into two parts of which one is the geometric mean of the whole and the other. This celebrated section is called the median or golden section, and the radius (or any tract so divided) is said to be divided in extreme and mean ratio. The problem of inscribing the lo-side is reduced then to the following : 264. Problem XXVII. — To divide a tract in extreme and mean ratio. Solution. Let a be the tract, and b the greater part ; then a\b\\b\a — b (Fig. 162), or a\ a -{- b \ \ b : a (why make this change ?) . Fig. 162. CONSTRUCTIONS. 203 Hence we may conceive of a- as the power of a point whose distances from the circle (along a diameter) are a-\- b and b (why?), so that a is the diameter of the circle. Hence, on a as diameter draw a circle ; at any point Z" of the circle draw a tangent and on it take TP= a ; draw the diameter PBA; then FB = b (why?), and the arc about F with radius b divides FT or a at I in extreme and mean ratio. Q. E. F. N.B. To the point /corresponds the harmonic conjugate O, the point of ou/er median section, such that F/:/T::FO: OT. Let the student show that FT- TO = pa, just as FT- TI^ F~l\ How shall we now construct a regular 5 -side, 20-side, • • • 5-2"-side? By combining the constructions for a 3-side and a 5 -side we may now construct a regular 15-side. For the difference- of the arcs subtended by a side of a regular 3-side and a side of a regular 5-side, is (|^ — 1^) of a circle, or ^-^ of a circle ; half of it is y^^, or (J — -^-^) of a circle, that is, the arc subtended by one side of a regular 15-side. Hence solve Problem XXVIII. — To cojistruct a regular \^'2'^-side. 265. At this point the query seems to arise naturally : if we can find the arc of a side of a 15-side by combining those of a 3-side and a 5-side, may we not find arcs of sides of other regular ;/-sides by other combinations? To take the most general case, let us form the difference of / arcs of a 2''«3-side and q arcs of a 2'-5-side; it will be the /~^ '^ )th of a full angle; i.e. it will be (2-5 — 2-3) 204 GEOMETRY. times the arc of one side of a i5-2'' + *-side ; but this latter polygon may be constructed by the preceding problem. Hence nothing new is obtained by the new combination. Herewith, then, the round of elementary construction of regular polygons is practically completed in the four series : 2'*-sides, 3-2"-sides, 5-2"-sides, i5-2"-sides. The profound analysis of Gauss has indeed shown that ruler and compasses will suffice to construct a (2'*+ 1) -side whenever (2" 4-1) is a prime number ; and accordingly we can construct regu- lar 17-sides (;z = 4) and 2 5 7-sides (;/ = 8) ; but the con- struction of the former is exceedingly tedious, and that of the latter is excessively so, while for still higher values of n the tedium and difficulty surpass all limit. However, in figures 94, 95 a regular 7-side and a regular 9-side are con- structed once for all, empirically, but to practical perfection. 266. Problem XXIX. — To draw a circle through two given points, tangent to a given ray. Hint. Consider the power of the intersection of the given ray and the ray through the points with respect to the required circle, and use Problem HI. 267. Problem XXX. — To draiv a circle through a given point and tangent to tivo given rays. Hint. Find a second point on the circle and apply Prob- lem XXIX. 268. The doctrine of perspective similarity may often be used in constructions. A. When one datu7n is a tract, the other data being angular and proportional relations. We then construct in accordance with these latter, disregarding the first one ; in the constructed figure a tract will correspond to the given CONSTR UCTIONS. 205 tract, and on this latter we then construct the required figure similar to the one first constructed. 269. Problem XXXI. — Given the angles and an altitude of a A, to construct it (Fig. 163). Fig. 163. Solution. Draw any A with the given angles ; then firom the proper vertex, C, lay off the given altitude as CD normal to the opposite side AB. Draw through Z) a II to AB cut- ting the other side at A^B\ Then A'B'C is the required A (why?). The two A ABC and A'B'C are perspectively similar, C being the centre of similitude. 270. B. IVken one figure is to be inscribed in another so that certain points of the one fall on certain lines of the other, we may draw a figure in perspective, with the required figure, as to the intersection of two rays on which are to lie two points, and then from this centre of similitude construct the required figure according to the remaining conditions. 271 . Problem XXXII. — To inscribe a square in a A 7vith the vertices of the square on the sides of the A (Fig. 164). Solution. Inscribe any square HP in the A, and draw ^/^ meeting BC at P' ) then /^ is a point of the required square. Complete the construction. 206 272. C. It is often required to inscribe in a given figure a tract that shall be cut by a given point proportionally in a given ratio. We may then assume the given point as centre of simiHtude, construct a figure similar to the given figure with the given ratio of simihtude ; then the required tract will go through a point of intersection of the two figures. 273. Problem XXXIII. — To draiv through a point I in a circle S {of radius r) a chord that shall be divided by I in the ratio a : b (Fig. 165). -.P \S' Solution. From / lay off opposite to IC the tract IC so that a : b : : IC : IC. About C as centre with radius r', such that a\b\\r\r\ draw a circle S meeting 6" at M and P. Then MN ox FQ is the chord sought (why?). CONS TR UC TIONS. 207 How will you proceed in case of outer division? The two divisions, inner and outer, may be conveniently distin- guished by prefixing the sign — to the smaller term of the ratio ; i.e. to the tract corresponding to the tract that will be wholly without the given tract after division. *274. The following discussions might have been intro- duced much earlier, at Miscellaneous Applications, but for interrupting the course of thought. We may conceive the area of a parallelogram as generated by slipping one of its sides along the other two parallel sides. Plainly, the side slipped is merely slipped or pushed, not turned at all, being kept parallel to itself (as the phrase is) throughout. Thus, suppose the tract AB slipped along the II and equal tracts AD and ^C; it will generate the paral- lelogram area ABCD. Clearly, the tract may be slipped along the same parallels in either of two opposite senses, as from A \.o D ox from D to A. The sense of the motion of the tract will be the same Fig. i66. 208 GEOMETRY. [Th. CXXX. as the sense of the motion of any point of its ray, as of /, the intersection of the ray with any other ray, as with the normal ray L (Fig. i66). The two senses of /'s motion may be distinguished as positive and negative ; then the cor- responding areas generated by the moving tract may also be distinguished as positive and negative. In summing such areas we always regard the sense and remember that to add resp. subtract a magnitude is the same as to subtract resp. add the counter magnitude., i.e. the magnitude equal in size but opposite in sense. Bearing this in mind we may now enounce : *275. Theorem CXXX. — The suin of the areas generated in simply slipping a tract round a A is o. Data: ABC the A, A A' the tract in its initial position (Fig. 167). C Fig. 167. Proof. Suppose the tract to compass the A counter- clockwise ; then if the area AB' be considered positive, the areas BC and CA' must be considered negative (why?). But on taking away the A A'B'C from the whole figure AB B'CA' there is left AB' ; and on taking away ABC there Th. CXXXIL] constructions. 209 is left the sum of BO and CA^ ; hence the areas AB^ and BC -{■ CA' are equal in size but opposite in sense ; hence their sum is o, or AB' -\- B C -\- CA' = o. q. e. d. Corollary. If the A ABC be curvilinear instead of recti- linear, the theorem still holds. *276. Theorem CXXXI. — If a tract be simply pushed round any closed figure, the sum of the areas generated will /^^o (Fig. 1 68). Fig. i68. Proof. The figure may be cut up into a number of A rectilinear and curvilinear. The sum of areas generated in compassing each A is, by the foregoing theorem, o ; hence the total sum of areas generated is o ; but each dividing tract, as AC, is compassed twice, in opposite senses, from C to ^ and from A X.o C ; hence the sum of areas gener- ated along these divisions is o ; subtracting which we have left the sum of areas generated along the outer border equal to o. Q. E. D. *277. Theorem CXXXII (of Pappus, a.d. 300) . — A par- allelogram 071 one side of a i\ ivhose counter-vertex lies between two parallel sides of the parallelogram, equals the sum of two parallelograms, on the other sides, whose parallel sides go through the vertices of the first parallelogra^n. 210 GEOMETRY. [Th. CXXXII. Proof. Let the student show from the figure, by help of Theorem CXXX that AB' = BC + CA (Fig. 169). Fig. 169. Corollary. As a special case, let the student prove the Pythagorean Theorem. 278. Def. The tract from a fixed point to a variable (or moving) point is called the radius vector of the moving point with respect to the fixed point. Thus OP is the radius vector as to O of the point P as it traces the curve C (Fig. 170). Fig. 170. Def. The area bounded by the path of the moving point and two positions of its radius vector is said to be generated, Th. CXXXIIL] constructions. 211 descri/?ed, or swept out by the radius vector in passing from the initial to the final position. Thus the area POQ is swept out by r in passing from OF to position OQ. Clearly, the same area may be described in either of two opposite senses, according as the rotation of the radius vector is clockwise or counter-clockwise, and the area must be distinguished accordingly. We shall call areas generated clockwise negative, and areas generated counter-clockwise positive. Remembering the laws for adding and subtracting magnitudes opposite in sense, we now enounce : 279. Theorem CXXXIII. — T/ie total area generated by a radius vector whose end compasses a A completely is the A itself. Proof. If the point O be within or on the A, the validity of the theorem is immediately evident. If the point O be (Fig. 171) without the A, then the area inside is generated but once, while the area outside, as AOC, is generated twice, in opposite senses ; once, as AOC, clockwise, once, as COA, counter-clockwise : such will always be the case, since the final and initial positions of the radius vector are the same. 212 GEOMETRY. [Th. CXXXIV. Hence the outside areas annul each other, and there is left only the inside area, the A. q.e.d. Corollary. If the A be curvilinear, the theorem still holds. 280. Theorem CXXXIV. — The area described by a radius vector whose end compasses any closed figure is the area oj the figure itself. Proof. Employ the method and reasoning of Theorem CXXXI. Conduct the proof carefully in the case of a ring and of a loop. Why do the arrows point as they do ? What effect will reversing one have on the other? Imagine the ring slit through from outer to inner border (Fig. 172). Fig, 172. The foregoing theorems play an important role in Higher Mathematics. THE TACTION PROBLEM. 281. In the following discussions certain higher concepts of Geometry, which have thus far been lightly passed over or not formed at all, become regulative and must receive graver consideration. We begin by re-defining some of THE TACTION PROBLEM. 213 them and recalling some of their already demonstrated properties. 1. The rectangle of the distances of a point from a circle along any ray through the point is called the power of the point as to the circle. The power is equal to the square on the tan gent- length from the point to the circle when the point is without, and equal to the square on half the shortest chord through the point when the point is within the circle. 2. All points that have equal powers as to two circles lie on a ray called the power-axis of the two circles. The ray is normal to the centre-tract of the circles, of radii r and /, and divides it into segments d and d^ such that {r+ r')-{r- r')z={d+ d'){d- d'). The three power- axes of three circles, taken in pairs, concur in a point called the power-centre of the three circles. 3. Any two points P and /" on the same ray through a fixed point O are said to be in perspective or perspectively similar as to the centre of similitude O in the ratio of similitude 0P\ 0P\ 4. Two figures are said to be in perspective ox perspectively similar when every point of one is perspectively similar to the corresponding point of the other as to the same centre and in the same ratio of similitude. 5. When OP and OP' have the same sense, the perspec- tive is direct, and the centre outer ; when they are opposite in sense, the perspective is counter, and the centre inner. 6. Any two circles are perspectively similar in the ratio of their radii as to both an inner and an outer centre ; namely, the points dividing the centre tract harmonically in the ratio of the radii, which are also the points of intersection of common tangents to the two circles, when such tangents there are. 214 GEOMETR. [Th. CXXXV. 282. Theorem CXXXV. — Lemma. — If two figures are similar to a third, they are similar to each other. The easy proof is left to the student. 283. Theorem CXXXVI. — If 07ie figure is in perspective with each of two, these latter are in perspective with each other, and the three centres of similitude are coUinear. Proof. Let P, Q, R hQ any three points in the first figure, F', Q, R^ and P^\ Q", R" the corresponding points in the other figures. Then RQR and P"Q"R" (Fig. 173) are similar (why?), and FR", QQ", RR" meet in a point, O' Fig. 173. th. cxxxvil] the taction problem. 215 (why?), as to which they are in perspective (why?). Like- wise PQR and P'QR' are similar, and PP\ QQ , RR' meet in a point 6?". Now let PQ, PQ, /^"(2" cut a O' at S, S\ S", and draw SR, SR', SR". Then QRS, Q'R'S', Q"R"S" are all similar (why?), 5 and S' are in perspective as to O", S and S" are in perspective as to O', and hence S' and S" are in perspective as to O. Hence O is on the ray O'O" (why?). Q.E.D. Corollary. Show that the three centres of similitude are either all outer or else one outer and tivo inner. Def. If a point bisect every chord of a figure drawn through the point, it is called the centre of the figure, and the figure itself is said to be centric. A central ray, and often a central chord, of the figure is called a diameter. 284. Theorem CXXXVII. — If tivo similar centric figures be in perspective as to one pointy -they are also in perspective as to a second point (Fig. 174). 216 GEOMETRY. [Th. CXXXVII. Proof. Draw two diameters of the one figure and the two corresponding chords of the other; these latter will also be diameters (why?), and the centres will correspond. Join the ends of these diameters crosswise; that is, the end of one with the non-corresponding end of the other. The intersection of these two cross-joins, /, is a second centre of similitude (why ?) . q. e. d. Corollary i. Of these two centres of similitude, the one is outer, the other inner ; and they divide the centre tract CC harmonically. Corollary 2. If three similar centric figures be in per- spective they have six centres of simiHtude, and of these the three outer are coUinear, as are also any one outer and the two other inner. Def. A ray on which lie three centres of similitude is called an axis of similitude. Corollary. There are four such axes, one outer and three inner. The central figure with which we have especially to do is the circle. 285. Def. When the rectangle of the distances from a fixed point O of two points, F and P, on the same ray through O, is constant, the two points are said to be inverse, or in inversion, with respect to O as centre of inversion. Def. A circle about the centre of inversion, with the side of the square equal to the rectangle of the distances for radius, is called the circle of inversion, and its radius the radius of inversion. Def. If while one of the inverse points as F describes a curve C the other describes a curve C, then C and C are said to be inverse or in inversion with respect to O. Th. CXXXVIIL] THE TACTION PROBLEM. 217 *ej. Let a ray through a centre of similitude of two cir- cles cut each in a pair of correspondent points P and Q^ P and Q ; then each point of each pair has a correspondent or homologous point in the other pair, as P and P\ Q and Q ; also each point of each pair has a non-correspondent, or contra-correspondent, or anti-homologous point in the other pair, as P and Q\ Q and P . 286. Theorem CXXXVIII. — Anti-homologous points of two circles are ifiverse with respect to the centre of similitude of the circles (Fig. 175). Fig. 175. Proof. Let O be the centre of simihtude of the circles, CC the centre ray. Then 2^ OAP= ^ O'A'P (why?) and ^ OAP= ^ BQP (why?); hence :^ BQP=^ BA'P ; hence :^ BA'P and ^BQP are supplemental; hence BA'PQ is an encyclic quadrangle ; hence OQ • OP = OB • OA'. Now the points O, B, A' are fixed ; hence the rectangle OB • OA' is con- 218 GEOMETRY. [Th. CXXXIV« stant ; hence Q and P are inverse as to O. Similarly prove that /'and Q are inverse, q. e. d. Corollary. OA • OB' = OT- OT ; hence the radius of inversion about (9 is the geometric mean of (^T'and OT', the tangent-lengths from the centre of similitude to the circles. 287. Theorem CXXXIV". — The inverse of a circle is i Is elf a circle. Data : In the figure let O be the centre of inversion, 6" the circle, / the circle of inversion with radius r. Proof. Draw 6^7" tangent to S and construct OT^ so that OT:r.'.r': 02\ Draw a normal to OT at T meeting OC Fig. 176. Th. CXXXVP.] the taction problem. 219 at C ; with radius CV draw a circle. It is the inverse sought. For it is the circle S of the preceding theorem (why?) J in which /*and Q, ^and P were inverse as to O. 288. Theorem CXXXV. — The transverse Joins {chords) of two pairs of anti-homologous points of two circles meet on the power-axis of the circles. Data : P and Q, V and U\ two pairs of anti-homologous points in S and S ; PF said Q'U', their transverse joins (chords) (Fig. 176). Proof. The quadrangle PFU'Q' is encyclic (why?). Let the student complete the proof. Def If a circle touch two other circles, the ray through the points of touch is called the chord (or, better, the ray) of Contact. 289. Theorem CXXXVI''. — The ray of contact of two circles with a third goes through a centre of similitude of the two circles (Fig. 177). Proof. For a point of contact of two circles is a centre of similitude of the two (why?) ; hence the ray of contact goes through two centres of similitudes ; hence it is an axis of simiHtude and goes through a third centre of similitude ; namely, of the two circles (why?), q.e.d. Corollary. When the two circles are touched similarfyy both innerly or both outerly, the ray of contact goes through the outer centre of similitude of the two ; when they are touched dissimilarly, one innerly the other outerly, the ray of contact goes through the imter centre of similitude of the two (why?). 290. Def The ray through one of two inverse points normal to their junction-ray is called the polar of the other 220 GEOMETRY. [Th. CXXXVII*. point, and this latter point is called the pole of the polar, with respect to the circle of inversion. This circle we may call the circle of reference, or the referee-circle, or simply the referee. Note carefully that pole and polar have no mean- ing except with respect to some referee. Fig. 177. 291. Theorem CXXXVII". — If of two points the first is on the polar of the second, then the seco?id is on the polar of the first. Data : S the circle, P the first point on the polar, Z, of the second point Q (Fig. 178). Proof. Draw the centre-ray OQ cutting L at Q, then Q is the inverse of Q (why?) ; also, let P be the inverse of P. Then the quadrangle PPQQ is encyclic (why?) ; also th. cxxxvip.] the taction problem. 221 the ^ ^' is a right angle (why?) ; hence so is the ^ /*' (why?) ; hence P^ Q is the polar oi P (why?), i.e. the polar of /* goes through 5. q.e.d. Fig. 178. Corollary i. The poles of all rays through P lie on the polar of/*; in other words, as a polar turns about a point, its pole glides along the polar of that point. Corollary 2. The polars of all points on a ray pass through the pole of that ray ; in other words, as a pole glides along a ray, its polar turns about the pole at that ray. Scholium. By definition the rectangle OP- OP^ is con- stant in area ; hence as P moves in towards O, P' moves out, and with it the polar of/*; as /* falls on 6>, P and the polar of P through it move out and vanish in infinity ; as P moves out from O leftward, P and the polar reappear in infinity on the left, approaching S , 2i?> P reaches S, so does P, and the polar becomes a tangent. Hence we may define the tangent as a polar whose pole is on it (the polar). 222 GEOMETRY. [Th. CXXXVIIl' 292. Theorem CXXXVIII'. — 7;z;/^, ^ is a fourth (part) oi F. The symbol for the w-fold of a magnitude is formed by prefixing m to the symbol for the magnitude. Thus, 2 A^ 3C, 40 , f7iM. The symbol for an ;;^th part of a magnitude is commonly formed by writing m below the magnitude and separating the two by a horizontal or oblique bar : thus, ^, ^, n/4, Qlm. 2 3 300. Theorem CXXXIX. — The p-fold of the mth part of a magnitude equals the mth part of the p-fold of the magni- tude. Proof. Let Q be any magnitude (tract, angle, area). By definition there are m ;;/th parts of it. In its /-fold each such part will be present / times ; hence there will be pm m\\v parts in the /-fold of Q. Also by definition the ;;/th part of this /-fold taken m times in summation must yield the whole. Now, however, if we take / of the wth parts of Q and take them m times in summation, we shall get a whole consisting of mpmih. parts. But it is a fundamental law of counting, called the Commu- tative Law of Multiplication, that to count in p times yields the same number as to count / ;// times. Hence this whole is equal to the /-fold of Q ; and its m\h part consists of / mi\i parts of Q, or is the /-fold of the mth. part of Q] i.e. the /-fold of the mth. part of Q equals the ;//th part of the /-fold of Q. Q. E. D, 228 GEOMETRY. [Th. CXXXIX. Scholium. The /-fold of the wth part, or the ;«th part of the /-fold, of Q is commonly written PQ P ^ Q — , or ' 'Q ox p ' — m^ m ^ ^ m The expression t., or p/ut is called a fraction, / and w its terms, p the numef-ator, f?t the denominator. We have just learned what it means. 301. If now we conceive any whole, w, as the sum of m equal parts, each equal to u, we may call u the unit magni- tude or magnitudinal unit. Thus one yard is a linear, one degree an angular, one acre an areal, unit. There may be several other magnitudes, the /-fold, ^-fold, jc-fold of this same unit u. Then w, /, q, x are called the metric numbers of these magnitudes. 302. It may happen that a magnitude may not be com- posable out of equal units u ; it may not be a multiple of the unit-magnitude u, but may be greater than the /-fold of u and less than the (/ 4- i)-fold of 11. Thus a circle is more than triple, yet less than quadruple, its diameter. In such cases it may be possible to find some smaller unit of which the unit u is the w-fold, and the other magnitude (say) the ^-fold. Thus the table may be more than 3 feet and less than 4 feet long ; but on changing the linear unit from the foot to the inch, the twelfth of a foot, we may find that the length in question is precisely the 40-fold of the new unit — the table is precisely 40 inches long. Then 40 is the metric number of the table-length in inches and the fraction ^.o. is the metric number of the same length in feet, which means that the sum of 40 12th parts of a foot is the length of the table. th. cxl.j metric geometry. ll") 303. Often, however, in fact generally, it will be impossi- ble to find any unit- magnitude so small that its w-fold shall be the one magnitude and its /-fold the other; and this im- possibility may be objective, not subjective — it may inhere in the nature of the case and not arise from some defect of our own powers of measurement or calculation. Thus, there is no unit-length, however small, out of which may be com- posed both the side and the diagonal of a square ; there is no length so small that the side shall be its w-fold and the diagonal its ^-fold. This important fact may be estabhshed thus : 304. Lemma. — If each of two tracts is a multiple of the same tract, the difference of the two is a multiple of the same tract. Proof. Let G be the greater and L the less of the two tracts. Then we have G=pt and L^q- 1; the difference of these two is (/ — q)t, and this is a multiple of /, since the difference of two integers,/ and q, is itself an integer,/—^. 305. Theorem CXL. — The side and diagonal of a square are incommensurable (Fig. 182). Proof. Let A^Bi = Si be the side, and A1A2 = di be the diagonal of a square. On A1A2 lay off A^B., = Si ; then A2B2 is the difference of the side and diagonal and is therefore a multiple of any tract of which Si and di are multiples ; call it .f2. Draw ^0^3 normal to A1A2 ; then B^A^ = ^^2^3 (why?) = A2B2 (why ?) . Hence ^2^3 is the diagonal, dz, of a square whose side is A^Bo or S2. Also ^2 is a multiple of any tract of which i-] and S2 are multiples. Hence we have a new square with side j-j, and diagonal ^2> both multiples of any tract of which s^ and di are multiples. Also the new side and diagonal are respectively less than half of the old side and diagonal. By repeating this process we obtain a third 230 GEOMETRY. [Th. CXL. square with side and diagonal less than half the side and diagonal of the second, less than one fourth those of Fig. 182. the first. By such constructions we shall obtain a square with side and diagonal less than {— Vh of the side and diagonal of the original square ; i.e. 2" 2^ By making n large enough we may make i"„ and d^, as small as we please, small at will, — we may in fact sink them below any assigned degree of parvitude. Meanwhile they must remain multiples of any tract of which s^ and ^1 are multiples ; but they can be multiples only of a tract smaller than themselves, manifestly ; hence any tract of which both Sy and dx are multiples must be smaller than a tract as small as we please. But there is no such tract, self-evidently. Hence there is no tract of which both diagonal and side of a square are multiples, q. e. d. Th. CXLI.] metric geometry. 231 Magnitudes that are thus not multiples of the same mag- nitude are said to have no common measure, or to be in- commensurable. 306. Now suppose the side s cut up into some very large number of equal parts, as q ; then the diagonal ^ will not be the sum of any number of these parts but will be more than the sum of/ parts and less than the sum of (/ + i) parts ; / /+ I that is, - ' s < d<, • s. q ^ q Here we may make q as large as we please ; hence, if we take s for our linear unit, we may shut in the metric number of ^ between two fractions, ^ and ^lJL^, that differ by -, q q q that is, by a fraction small at will, while the metric number of d differs from each by less than -. In this last sentence q we have subreptively assumed that d has some metric number when referred to s as unit-length. But we shall not build anything on this assumption at present. See Art. 256. 307. We now pass to the metric numbers of area. We agree once for all that the square on a linear unit shall be an areal unit. The metric number of an area will then be the number of areal units of which it is composed, or its equal is composed. 308. Theorem CXLI. — The metric number of a rectan- gle is the product of the metj'ic numbers of its dimensions. Two cases arise : T. When the dimensions are conunensurable (Fig. 183). Let a and b be the dimensions of a rectangle. Choose any unit of which a and b are both multiples, as u, so that 232 GEOMETRY. [Th. CXLI. a= p • u, b = q ' ti. Then we may cut up a into/ parts, and b into q parts equal to u. Through the points of division draw Hi" to the sides. Then the whole rectangle will be cut n h Fig. 183. up into pq squares each on the side u (why?) ; hence the rectangle will be the sum of these areal units ; hence the metric number is the product/^, q.e.d. Now suppose we choose some larger areal unit, as the square not on u but on 7^u. In this square there are, by the foregoing, rr units ; that is, it is the rr-fold of the small unit w ; or the small unit is the ( - Jth part of the large areal unit ; hence the rectangle, which is the pq-ioXd of the small (pq\ unit, is the — th of the large areal unit. But the sides ^"^ (P\ fA were respectively the ( - 1th and the ( - jth of the small linear unit u ; and the product of these two fractions is, by the Pq laws f 07- i7iultiplying fractions, — . Hence, if we call frac- tions numbers, we must have the metric number of the area is pq ^, or is the product of the metric numbers of the dimen- sions. Q. E. D. 2. Now suppose the two dimensions incommensurable with the linear unit u. Then a will be > /// and < (/ + i ) // ; Th. CXLL] metric geometry. 233 while b will be ^ q • u and < (^ + i)?^. Then plainly there are pq squares on // in the rectangle, with some remainder, but not {pq -\-p 4-^+1) squares ; or pq . ?/ - and - but < r r r q -{- 1 and , and the metric number of the area, if it have a pq (/+ i) (^+1) metric number, will be > — but < : i.e. it ' rr rr ' will lie between these two fractions and differ from each by / + $^+ I less than . Now with any fixed unit length, as i, P ^ we may find two fractions, - and -, that differ from the metric numbers of a and b (if they have any) by less than -> and by taking r even greater and greater we may approach p q our fractions - and - close at will to the metric numbers of r r p q a and b. Each of these fractions - and - meanwhile remains r r less than some assignable whole number ; so, too, does their p -\- q p -{- q -{- 1 sum , and so does . Now the difference of r ' r the fractions — and is the rth part of rr rr ^ , and by making r large at will we may make this 234 GEOMETRY. [Th. CXLII. rth part small at will. Hence the two fractions may be brought as close together in value as we please, while between them lies always the metric number of the area, and also between them lies always the product of the metric numbers of the dimensions. These two numerics, then, the metric number of the area and the product of the metric numbers of the dimensions, cannot differ by any assignable value however small, since they both He between two values which may be made to differ by less than any assigned value however small. Hence we conclude (ist) that these two numerics are Definites, since the bounds between which each lies, and which close down together upon each other, are at every stage perfectly definite, and (2d) that they are abso- lutely the same in value. 309. It is a matter not of logical compulsion but of con- venient choice to call this Definite a number or at least a numeric. Since it is not expressible as a fraction, still less an integer, it is commonly called an Irrational. The laws of operation on the algebraic symbols of such Irrationals as well as Fractions are not matters of logical proof, but of allowable assumption. It is convenient to assume for them, arbitrarily to impose upon them, the same laws of operation that are found empirically to hold for positive integers, or numbers obtained by counting. This fact is sometimes called the Principle of the Permanence of the Formal Laws of Operation (Hankel). Further discussion of the subject belongs to Algebra and would be out of place here. 310. Knowing that the metric number of a rectangle is the product of the metric numbers of its dimensions, we now declare at once that Theorem CXLII. — The metric number of a parallelogram is the product of the metric numbers of its dimensions. Th. CXLIV.] metric geometry. 235 Theorem CXLIII. — The metric number of a A is half the product of the metric numbers of its di?nensiojts. Theorem CXLIV. — The metric nurtiber of a trapezoid is half the product of the metric numbers of its altitude and the sum of its II bases. In a word, all the theorems that declare relations among areas may now be translated into theorems that declare like relations among the metric numbers of areas. This easy exercise is left for the student. 311. We have thus far treated proportion strictly geo- metrically. We have written off the symbolism a\b\\c\d, when a^ b, c, d, were signs for tracts, but when asked what we meant by it our only reply was, we mean that the rectangle ad equals the rectangle be. This reply was perfect and complete. Now, however, if instead of the tracts we put p, q, r, s, as the metric numbers of the tracts, we may still write as before p : q : '. r : s, and answer the question what this means, by saying it means that the product ps equals the product qr, for we have just proved this equaHty. This answer is also perfect and com- plete. However, it is not the only possible answer. For we might say we mean that the fraction^ equals the fraction - ; q s this would also be correct. For if ps = qr, then on divid- ing both sides by qs we get - = - , and conversely, if ^ — _, q s ^ ^ then on multiplying both sides by qs we get ps = qr. Ac- cordingly these two answers are equally adequate and involve 236 GEOMETRY. each other. But we could not make any such second answer to the question, what do we mean by the proportion a: b:: c : d} For we cannot attach any meaning to the symbolism - when b a and b are tracts, nor can we tell, at least at present, what we mean by dividing one tract by another. We may indeed write - = - , and answer the inquiry as to our mean- b d ing by saying we mean the rectangle ad equals the rectangle be ; but we cannot deduce the relation reetangle ad = rec- tangle be from the symbolism - = - by multiplying through b d by the rectangle bd, for we do not attach any meaning to the phrase " multiplying by a rectangle." 312. The state of the case then is this : All the proportions among tracts in Geometry may be supplaced by corresponding proportions among the metric numbers of those tracts ; in these latter proportions we may supplace the colon by the division—, or quotient—, or fraction-mark, and the double colon by the equality-mark. The ratio of two tracts we did not attempt, and did not need, to define geometrically ; but we now define the corre- sponding ratio of the metric niitnbers of those tracts as the quotient of the one metric number divided by the other ; and a proportion among these metric numbers of tracts we may define as an equality of ratios. 313. We may now boldly apply the ordinary laws of algebraic equations to any geometric proportion, understand- ing by its terms not the tracts themselves, but the metric numbers of the tracts. The result will be some relation among the metric numbers of tracts. If desirable, we may Th. CXLIII".] METRIC GEOMETRY. 237 at once translate this relation back into pure Geometry by substituting for the metric numbers of the tracts the tracts themselves. But it will not always be possible to interpret geometrically the result of this substitution. An illustration will make this clear. 314. Let/, r, s, t, u be the metric numbers of the tracts on which they are written in the A ABC, right-angled at B with BD normal to ^C (Fig. 184). Then /= fu (why?), and /= /V= (^2- j^) (/- J-2) ^fr'-fs- Whence /V^ =/V _^ r^^^ whence + .-4-rV 5^ p-"^ r" This beautiful and important relation may be stated thus : Fig. 184. Theorem CXLIII". — The 7-eciprocal of the squared metric number of the altitude to the hypotenuse of a right triangle equals the sum of the reciprocals of the squared metric Clum- bers of the sides. So stated the meaning is intelligible and unmistakable. But if now we write for s, r,p the tracts themselves, namely, BL)- AB^ BC"' then we may indeed understand this relation algebraically precisely as before, meaning by the signs BD , AB , BC 238 GEOMETRY. [Th. CXLIIP the squared metric numbers of the tracts BD^ AB, BC ; but we cannot attach any geometric meaning to the equation, for we cannot tell what we mean by the reciprocal of a geo- metric square. 315. The next illustration is still more interesting and important (Fig. 185). Fig. 185. Let ABC be any A, L any ray cutting the sides at A\ B\ C ; from A, B, C drop normals on Z meeting it at B, Q, R. Then by similar A we have AC.BC.-.AF'.BQ, BA^ : CA' -..BQ.CR, CB' :AB^ ::CB: AP. If now we understand by these biliterals not the tracts themselves, but the metric numbers of the tracts, the fore- going proportions will still hold and may be read as equa- tions and written thus : AjC^AP BA[^BQ CB' ^ CR BC BQ' CA' CR' AB' AP' Th. CXLIV«.] metric geometry. 239 where the sides of the equations are ordinary fractions. On multiplying them together there results BC^'CA^'AB' '^^''^' Inasmuch as ^C and BC^ are reckoned oppositely as are also BA^ and CA\ BC^ and AB\ it is common and convenient to write — i instead of i, thus : AC ' -BA'- CB ' ^ _ J BC'CA'-AB' This is the celebrated proposition of Menelaos : Theorem CXLIV". — The continued product of the ratios in which a ray cut the sides of a A is — i . It states the condition necessary and sufficient that three points on the sides of a A shall be collinear, and its mean- ing is perfectly clear so long as we mean by ^C, etc., not the tracts but the metric numbers of the tracts. But in order to interpret it geometrically, AC\ etc., standing for the tracts themselves, it would be necessary to define pre- cisely a higher notion, namely, that of the volutfte of the cuboid of three tracts, and this would require us to pass out of our plane into tri-dimensional space. 316. Still another illustration is found in a proposition the logical complement of the preceding (Fig. 186). Let any three rays through the vertices of a A concur in O, and let normals from O meet the sides of the A AB C at the points A\ B\ C. Draw OA, OB, OC, and form the pairs of ratios OA' : OB ; OA':OC; OB'-.OC; OB' : OA ; OC'.OA; OC : OB. Regarding them as ratios not of tracts but of the metric numbers of tracts, we may treat them as fractions and form 240 GEOMETRY. [Th. CXLV. the product of the first in each of the three couplets, and also of the second in each couplet; these products are evidently equal. Now the fraction —— depends for its LJA value solely on the angle a ; hence it is called a function of G the angle a, namely, the sine of the angle «, and so for the others. Hence sine of a sine of R sine of y . ■ ci . L — I • or sine of «' sine of ^S' sine of y' Theorem CXLV. — The cofttinued product of the ratios of the sines of the angles into which three concurrent rays through the vertices of a l\ divide the angles of a A is i. The converse of this theorem is easily established, which accordingly supphes a test of the concurrence of three rays through the three vertices of a A. Its meaning is perfectly precise and unmistakable so long as not the tracts but the metric numbers of the tracts are signified by OA, etc. ; otherwise we are not in position to prove it nor to interpret the symbolism expressing it. METRIC GEOMETRY. 241 317. The notion of sine of an angle^ introduced for sim- plicity in the foregoing article, is of the highest importance for all following geometrical study. But perhaps a more fundamental notion is that of cosine, which we may define thus : Def. The ratio of the projection of a tract on a ray to the tract itself is called the cosine of the angle between the ray and the tract (Fig. 187). ----'ar V Fig. 187. Thus the ratio of the projection / of the tract / to the tract itself is the cosine of the angle a between them ; or / : / = a, as we may write cosine of a, which is commonly abbreviated into cos a. 318. It is plain that the projections of t on parallel rays are all equal ; hence we may suppose the ray of projection drawn through the beginning of /, as in Fig. 188. Then as / turns about o and « changes its value, the projection p oi t will also change. Thus : for « = o, 60°, 90°, 120°, 180°, 270°, 360°, 420°, ••• «= I, h o, —1 —I, o, I, i ... 319. In the 2d and 3d quadrants the projection / is reckoned leftward ; it is opposite in sense to the projection when a is in the ist or 4th quadrants, and accordingly the cosine is marked — . When a increases by 360° (or 2 tt, see Art. 336) from any value, the revolving tract resumes its 242 GEOMETRY. original position, the projection resumes its original value, and so too does the cosine ; hence cos (360° + «)= cos (2 7r + «) = cos« j that is, the cosine is not changed by increasing (or decreas- ing) the angle by a round angle. Hence, plainly, cos(/)t ± 2 mi) = cos a. Hence the cosine is called a periodic function* of the angle, the period being 2 tt, that is, a round angle (see Art. 336)- 320. If /■ be turned through any angle a from the posi- tion OA, its projection/ is the same whether the turning be clockwise or counter-clockwise ; that is, the projection is the same whether the angle be negative or positive ; hence, too, the cosine is the same. That is, cos ( — «) = cos a ; that is, the cosine of the angle is unchanged by changing the sense (or sign) of the afigle. Now we learn in Algebra that only the even powers, not the odd powers, of a symbol are unchanged by changing the sign of the symbol ; thus : (_«)2^«2^ (^-ay=a\ but (- aY=-a\ Hence the cosine is called an even function of the angle. Its periodicity and its evenness are the two cardinal prop- erties of the cosine, on which all others hinge. 321. We may now arbitrarily define the sine of an angle to be the cosine of the complemental angle ; i.e. go° — a z= a \ , as we may write sine of a, which is commonly abbreviated into sin a. * Two magnitudes such that the values of the one correspond to values of the other are C2\\&d. functions of each other. METRIC GEOMETRY. 243 Now write 90° — /? for « ; we obtain cos \ 90° - (90° - ^) S = (90° _ ^) I, or ^ = (90° - ^) | ; i.e. the cosine of an angle is the sine of the complemental angle. Hence the sine (or cosine) of either of two complemen- tal angles, as « and y8, is the cosine (or sine) of the other ; i.e. \i a-\- ^— 90°, then « = /? |, and « | = ^. When either changes by 2 tt, so does the other oppositely ; hence the sine as well as the cosine returns into its original value ; i.e. the sine is also periodic with the period 2 tt. 322. If the tract /be reversed, that is, turned through a straight angle, its projection will also be reversed, but other- wise unchanged ; hence the cosine will be reversed ; that is, cos(M-l-7r) = — cos« j also cos(a — 7r)=cos(«-f-7r) (why?) ; hence cos (« — tt) = — cos «, or, to change the angle by the half-period, tt, changes the sign of the cosine. 323. Since the cosine is an even function, cos (« — tt) = cos (tt — «) ; hence cos (tt — «) = — cos a. That is, since a and tt — a are supplemental, the cosines of supplemental angles are counter — equal in size, but opposite in sense (or sign). 324. Again, if « -f /? = 90°, then ol\ — ^ (why ?) . Then (« + tt) | = ^-tt (why ?) = - ^ = - « | . That is, to change the angle by the half-period, tt, changes the sense of the sine as well as of the cosine. 244 GEOMETRY. P2 % ^ -^ 3 / \ V 1 / \ / ^ \ p.' \ /Pa R' Po 1 y^o ^^ i Po' 1/ K y >■ Fig. i88. Fig. 190. METRIC GEOMETRY. 245 We now ask, how is the sine affected by changing the sense of the angle ? We have (-«)| = 90° +« (why?) = - (9 0° - ci ) (why?) =-(«)!; that is, the sense of the sine changes when the sense of the angle changes. But this is the property only of odd powers, not of even powers, of a symbol ; thus ( — ay = — a^, {— aY = —a^, etc. Hence the sine is called odd function of the angle. Its periodicity and its oddness are the cardinal properties of the sine, on which all others hinge. 325. If q be the projector of the tract / then - is plainly the sine of the angle a for every position of t. We may indeed define the sine of a as equal to this ratio, and from this definition readily deduce all the foregoing properties (Figs. 188, 189, 190). Exercises, i. Prove that {a\y -\- (a)^= i. 2. Find the value of a \ for a = o, 30°, 45°, 60°, 90°, 1 20°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 330°, 360°, 390°. 3. If a, b, c, be the sides of a A, a, ji, y the opposite angles, ;• the circumradius, prove that (f' _ b _ c _ Such is the Law of Sines (Fig. 191). 4. Prove that a- = U"- -\- c' — 2 l)ca, — Law of Cosines. 5. If ^ and b are adjacent sides of a O, and ab denote the angle between them, prove that EJ = ab- ab\. N.B. We may define the sine from this important theorem thus : The sine of the angle betiveen two sides of a CD is thai 246 GEOMETRY. number which taken as a multiplier turns the area of the rectangle of the sides into the area of the O. Fig. 191. 326. If we project successively the sides of a closed poly- gon on any ray, the sum of the projections will be o ; for the end of the projection of the last side falls on the beginning of the projection of the first. This fact is very important in surveying, where in compassing a field the sum of the northings must equal the sum of the southings, and these two sums, having opposite senses, together make the whole sum o. So for the eastings and westings. We may express this fact in symbols thus : j-i^i + j-g^a + >y3«3 + • • • 4- s,,c(y, = 0= %sa. Here the j"'s are the sides, the a's are the angles of the sides with a fixed ray, as the east and west line, sa is the pro- jection of a side, and 2 is the symbol of summation. Exercise. Show by projecting on a ray normal to the first ray that ^sa \ = o. 327. If we project consecutively all the sides of a polygon but one on that one, the sum of the projections will be that one itself. For the beginning of that side is the projection METRIC GEOMETRY. 247 of the beginning of the first, and its end is the projection of the end of the last, of the projected sides (Fig. 192). Thus, in a A, the sum of the projections of two sides on the third is the third. 328. We make a most important appHcation of this simple fact in finding the cosine of the difference of two angles (Fig. 193). Fig. T93. 248 GEOMETRY. Let (9/* and OQ make angles a and ^ with any fixed ray OX. Then angle QOF = a — p. Now project any tract OQ on OP; there 0F= OQ- a — ^. But instead of pro- jecting OQ directly we shall obtain the same result by pro- jecting consecutively OF and FQ . The projection of OF is OF -a, and of FQ is FQ-a]. But 0F= OQ-jB, and Hence 0F=: OQ' a-(i = OQ-a^^^ 6>(2-«|-)8|. Or, «-/? = «. /? + «|.y8|. By changing the sense of ^ and by putting 90 — a mstead of a, let the student show that « + j8=a«)8 — a|-y8|, (« + /?)! = «|.^ + .^./?l, («-^)i=:«|.^-«r^i. These four formulae express the Addition-Theorem of Sine and Cosine. The doctrine of Functions of Angles constitutes Trigo- nometry — an extremely important subject, which cannot be pursued any further here. See Smith's Clew to Trigo- nometry. MEASUREMENT OF THE CIRCLE. 329. Thus far our linear measurements, or comparisons of length, have been wholly of tracts. The peculiar sim- plicity of such operations is due to the fact that any tract may be superposed (at least in thought) on any other, and thus their equality or inequality infallibly tested. We may similarly compare arcs of equal circles, but not arcs of unequal circles, nor arcs and tracts ; for these cannot be made to fit on each other to even the smallest extent. We feel sure indeed that a circle or arc has a perfectly definite length, that it is longer than some tracts, shorter than others, and equal to some others. For if we suppose an inextensi- MEASUREMENT OF THE CIRCLE. 249 ble cord wrapped around a circular disk, on unwinding and straightening the cord we should obtain a tract equal to the circle in length. But it remains difficult or impossible to fix the notion of the length or to determine the length itself without some preliminary definitions and assumptions. We assume then that a circle has a definite length, neither more nor less ; also, that it bounds a definite area, neither more nor less. 330. We now inscribe in the circle of radius r a regular ;^-side, and parallel to this latter we circumscribe a regular ;?-side ; then bisecting each arc subtended by a side of the inscribed ;^-side we inscribe a regular 2 «-side and also cir- cumscribe parallel to it a regular 2 /z-side. Then the following facts are at once evident : 1 . The area of any inscribed polygon is less, and the area of every circumscribed polygon is greater, than the area of the circle ; or if /„, S, and C„ designate these areas, then h/,„ C„,< C„ (why?). 3. The area of each regular «-side is half that of the rect- angle of the perimeter and the central normal on a side, and in case of the circumscribed polygons this normal is the radius r, but in case of the inscribed polygons this normal, or apothem^ a,^, is less than r. 4. The perimeters of inscribed and circumscribed ;?-sides are to each other as a^ and r ; for they are similar polygons, and «„ corresponds to r. 250 GEOMETRY. 5. Since the area of the circumscribed ;/-side decreases as n increases, and since one dimension, the radius, remains constant, it follows that the other dimension, the (half) perimeter, must decrease with increasing n. For n = 4 the polygon is a circumsquare, and the perimeter is 8 r. 6. Since the sum of two sides of a A is greater than the third side, it follows that the perimeter of the inscribed //-side increases with increasing n. For n = 6 the perime- ter is 6 ;-. Hence for all higher values of 7?. the perimeters of both inscribed and circumscribed polygons lie between 6 r and 8 r. 7. Since then the sum of the //-sides is certainly less than 8r, by making ;z large enough we can make each side, whether of inscribed or circumscribed polygon, as small as we please, smaller than one millionth, smaller than one bil- lionth, smaller than any assigned magnitude however small. 8. But the half-side of the regular inscribed //-side is a geometric mean between the segments of the normal diame- ter ; i.e. S As S^ is small at will, so is — , and still more is r — ^„, which we may call d^, the distance between the parallel sides of the inscribed and circumscribed polygons. N.B. A magnitude small at will is often called an infini- tesimal. Since /' — a^ or d^ is infinitesimal with respect to —J which is itself infinitesimal, it {d„) is called an infinites- imal of 2d order. But we are not now concerned with this fact. MEASUREMENT OF THE CIRCLE. 251 9. The difference in area between the circumscribed and the inscribed ;/-sides is a trapezoid the half-sum of whose parallel sides is less than ^r, the altitude being d^. Since 8 r is finite and definite while d^ is small at will, it follows that the difference in area of the circumscribed and inscribed regular polygons is small at will. Fig. 194. 10. Again, we have from similar homothetic figures, Perimeter of C„ : Perimeter of /„ : \r\r— d^. Hence, dividendo, Perimeter C^ : Perimeter C„ — Perimeter I,,\\r\ d^. But d^ is small at will ; so too then is /?„ = Perimeter C„ — Perimeter of /„ ; i.e. the difference in perimeter of the circumscribed and inscribed regular polygons is small at will. 1 1 . The circle lying always wholly between the two poly- gons inscribed and circumscribed both in position and in areal value, it follows that the difference between its area 252 GEOMETRY. and the area of either polygon is small at will ; and since the circle is fixed both in area and in position, it follows that the circle is the Limit of both polygons, and that the polygons close down upon it close at will as n grows ever larger and larger. 331. Think of the polygonal strip between the inscribed and circumscribed ;/-sides as growing ever narrower and narrower ; the circle is a fixed boundary toward which the circumscribed ;z-side shrinks down as n increases, and in such a way that there is no assignable point outside of the circle, no matter how close to it, that will not also fall out- side of the circumscribed /^-side as n increases. Likewise the circle is the fixed boundary toward which the inscribed «-side swells out as the ;/ increases, and in such fashion that there is no assignable point inside of the circle (no matter how close to it) that will not fall inside of the inscribed «-side as n increases. Thus the inscribed and circum- scribed /^-sides close down upon each other so as to leave no point between except the points of the ci7'cle itself. As n increases, the polygons tend to absolute coincidence with each other and with the circle. This fact is expressed fully and accurately by saying that the circle is the common Limit in length, in area, in position, of the inscribed and cir- cumscribed regular 7^-sides for increasing n; it is expressed elliptically and inaccurately, but conveniently and frequently, by saying that the circle is or may be regarded as a regular polygon of a 71 infinite number of sides. 332. The area of a circumscribed ;z-side is half the area of the rectangle of the radius of the circle and the perimeter of the polygon, for all values of n. Hence the area of the circle is half the rectangle of radius and the perimeter ; that MEASUREMENT OF ANGLES. 253 is, the circle. The numeric expressing the ratio of the circle to its diameter is called the perwietric ratio, and is desig- nated by the Greek initial of perimeter, tt. It is an irra- tional and hence not expressible exactly as a fraction, whether common or decimal, but its value has been calcu- lated in various ages to various degrees of exactitude, and may be calculated to any degree of exactitude. Of late years it has been calculated (by Shanks) to the 707th decimal place, and verified to the 500th — a degree of accuracy immensely higher than can be attained in any measurement. For most practical purposes the value tt = 3.14159 or even 3. 14 1 6 is close enough. If then r be the radius, 2 irr is the length of the circle, and 7rr • r, or ir/^, is its area. 333. Because the ratio tt is irrational it by no means fol- lows that it cannot be constructed geometrically ; that is, that we cannot with ruler and compasses draw a tract that shall be exactly equal to the circle of radius r. The ratio "\/2 is irrational, yet we can easily construct V2 • r, by drawing the diagonal of a square of side r. If we could draw a tract irr, equal to a half-circle, then, by Problem III^ p. 193, we could construct the geometric mean of r and nr, which would be the side of a square precisely equal to the circle in area. This famous problem of squiuing the circle is therefore not an irrational one ; it is unsolved, but possi- bly not in itself unsolvable. But see Math. Ann., xx., p. 213. MEASUREMENT OF ANGLES. 334. We have denoted by tt the so-called perimetric ratio, namely, of the circle to its diameter, 2 r, so that, if r be the radius, then 2 irr is the (length of the) circle, and ir/^ is its area. But there is another important use of ir. 254 GEOMETRY. [Th. CXLVI. 335. We have learned the ordinary sexagesimal division of the round angle into 360 equal parts called degrees. This artificial umt, degree, does not recommend itself for purposes of mathematical investigation, but a natural unit is suggested by the measurement of the circle itself. For it is plain that whatever part an arc is of the whole circle, that same part the central angle of the arc is of the round angle. Thus, if m times the arc make out the circle, then m times the central angle will make out the round angle ; for, in add- ing the arcs about the centre O, we at the same time add the angles at O subtended by the arcs. If, however, arc and circle be incommensurable, cut the arc and also the angle into q very small equal parts. Then / of these arc- parts will be less and / -f i will be greater than the circle, while, similarly, p of the angle-parts will be less and / + i will be greater than the round angle ; and this will always hold, no matter how great q and p may be. Hence, using the arc as unit-arc and the angle as unit-angle, we see that the metric numbers of circle and round angle lie always between the same fractions, - and ; and for increasing / and q these fractions close down upon each other, so that the metric numbers of circle and round angle cannot differ by ever so little, but must be precisely the same. If instead of circle and round angle we take any other arc and its corresponding central angle, the reasoning remains unchanged, so that we have Theorem CXLVI. — If any arc and its corresponding central angle be taken as units, the metric numbers of any other arc (of the same or equal circle) and its corresponding central angle are the same. In other words, THE EUCLIDIAN DOCTRINE. 255 Central angles and their subtending arcs (in the same or equal circle) aie proportional (see Art. 337). The latter statement is conciser ; the former is preciser. 336. Now a natural unit for arc-measurement is plainly radius ; hence a natural unit for angle-measurement is the angle whose arc is radiwi^ (in length) ; accordingly we adopt it as unit-angle and name it Radian. The metric number of the circle, radius being unit, is 2 tt ; hence the metric number of the round angle, radian being unit, is 2 tt. The radian equals about 57° 17' 7,7,^'. Corollary. If ;/ be the metric number of any angle, and therefore of its corresponding arc, radian and radius being units, then of any other arc, subtending the same or equal angle, but described with a radius whose metric number is r, the metric number will be nr ; for all circles are similar. That is. The metric number of an arc equals the product of the metric numbers of its central angle and its radius. Exercise. What are the natural metric numbers of a straight angle ? A right angle ? An angle of 60° ? Of 45° ? Of 30°? Of 120°? 150°? 225°? 240°? 270°? 420°? 600°? 720°? 1080°? THE EUCLIDIAN DOCTRINE OF PROPORTION. 337. According to Euclid, / nb, according as mc < 7td, mc — nd, or mc > nd. We may also say equivalently that a has the same ratio to b that c has to d when, afid only when, etc. 338. Hereby is defined, then, precisely, not indeed ratio, but at least equality of ratios. However, it still remains to be proved that the axiom of equal magnitudes, namely, mag- nitudes equal to the same or equal magnitudes, ai'e equal to eath other, can be applied to equal i^atios ;■ for it has not yet been shown that ratios are magnitudes or may be treated as magnitudes. The all-important fact that, whatever these ratios may be, they obey the axiom of magnitudes, is expressed in the Theorem CXI'\ — \i a \ b w c \ d and a\b\:e:f, then c : d : : e :/ (see Theorem CXI.). For herein is declared that when two ratios, c : d and e :/, are equal to the same ratio, a : b, they are equal to each other. After establishing this fundamental proposition, but not before, we may drop the double colon, : :, and write a '. b = c : d. 339. We may illustrate both the idea and the method of Euclid, in demonstrating the following extremely useful Theorem CXLVII. — Areas of similar figures are to each other as the squares on homologous tracts (in the figures). Th. CXLVIL] the EUCLIDIAN DOCTRINE. 257 . Data : F and F^ two similar figures, / and /' two homol- ogous tracts, /^ and f^ the squares upon them. Proof. If the figures be curvilinear, and in general even if they be rectihnear, it will not be possible to cut them up into corresponding squares, however small, as is manifest. Nevertheless, we may cut each one up into congruent squares so small that the remainder shall be less than any assigned area, however small ; that is, shall be small at will. For (Fig. 195) draw two corresponding series of equidistant horizontals and verticals, d apart in i% and d^ apart in F\ Fig. 195. Let the extreme verticals in /^ be ^ apart, and let there be e; + I of them, so that g—vd. Then the excess of the area of F over the sum of the squares will be the sum of the pairs of pieces at the end of each vertical strip ; hence it will be less than 2 v little squares, or less than a rectangle of base 2g and altitude d, i.e. < 2gd. But this rectangle is small at will, since its base 2 ^ is constant and finite while its alti- 258 GEOMETRY. [Th. CXLVII. tude d is small at will. Similarly, in F\ 2 ^V/' is small as we please. Now suppose there are p little squares cut out in F and also in F\ where / is some very large but perfectly definite integer. Then the areas A and A' of F and F' will differ from the sums of the squares, pd^ and/^'^, by s and s', two magnitudes small at will ; i.e. A=pd-'^s and A' = pd''' ^ s\ Now suppose md'^ = m^d^' ; i.e. the sum of ni squares in F equals the sum of ;;/ squares in F\ Plainly then p {md^) =p {m^d''') ; or m -pd^ = m^ -pd^"" (why?). Now mA = m ' pd^ + ms and m^A — w' -/^'^ + m}s^ ; hence mA — m'A' = ms — m^s\ But s and ^' are small as we please, while m and /«' are finite j hence wi- and m^s^ are small at will ; hence their difference, ms — m^s\ is less than a magnitude small at will ; hence it is o. (Why? Because there is only one definite magnitude, namely, zero, that is less than a magnitude small at will.) Hence, if i7id^ = wV/'^, then mA = 7n^A\ Now suppose md-^7n}d^'^. Then, as before, mpd^>m'pd^^j and mA = mpd^ -\- ms while m^A' = m^pd^^ + m^s\ Hence mA — m^A^ = {m -pd^ — //z' -Z^'^) -\- ms — m's\ Here again ms — m's' is small at will, while /// •pd'^ — m'-pd^' is some finite magnitude ; hence mA — mA^ is also some finite magnitude of the same sense ; i.e. if md^ > m'd^^, then w^ > m'A'. Precisely in like fashion we prove that if md'^ < 7n'd''^j then mA < mA'. Th. CXLVIP.] the EUCLIDIAN DOCTRINE. 259 That is, niA < m^A\ mA — m^A', or fnA < ffi'A\ according as md''7n^d'\ Hence, by definition, the areas are proportional to the squares on the corresponding tracts, d and ^' ; or A: And': d'K Now compare the squares on / and f', d and d'. Since all squares are similar, and since d and d' are corresponding tracts or the equals of corresponding tracts in the squares on / and /', we have from the foregoing. Hence, by the Axiom of Magnitudes, applicable to ratios, AlA'-.-./'-.r-. Q.E.D. 340. Care has been taken to conduct the foregoing demonstration so that it shall apply quite as well to circles, to regular triangles, in fact to any similar figures drawn on homologous tracts as to squares ; so that we may afiirm Theorem CXLVII". — Areas of sitnilar figures are pro- portional to areas of any similar figures on homologous tracts (in the original similar figures). 341. The peculiar propriety and advantage of using the square are seen on stating the analogous arithmetical Theorem CXLVII\ — The metric numbers of similar figures are proportional to the metric numbers of any other similar figures in the same ratio of similitude. Now the figure (or area) whose metric number is easiest to find is the square^ whose metric number is the second power of the metric number of its side. Instead of second 260 GEOMETRY. power it is usual, almost universal, to say square and thus employ this latter term in two entirely different senses, — the proper geometrical sense and the tropical arithmetical sense. This double use of the term " square " is very regrettable as being especially confusing to beginners. 342. We may now restate our proposition thus : The metric numbers of two similar figures are propor- tional to the metric numbers of squares on homologous tracts ; Or, are proportional to the second powers of the metric numbers of homologous tracts ; Or, are in the duplicate ratio of simihtude of the figures themselves. Thus, if the ratio of similitude be 2:5 ox a\b, the ratio of the areas will be 4 : 25 or ^- : ^l ^ 343. Observe carefully that the Euclidian doctrine of proportion is not a geometrical doctrine, but an arithmetical doctrine applied to Geometry. The same may be said of the accepted doctrine in modern texts : it is Arithmetic applied to Geometry. Nevertheless, the difference between the two is very great. Euclid's is based wholly on the oper- ation of multiplication and employs only positive integers, not fractions nor irrationals, which indeed the Greek did not recognize as numbers ; the modern, on the other hand, is based on the operation of division, and necessarily involves some general theory of fractions and irrationals. Moreover, Euclid's, while regarded as cumbrous and very difficult for the beginner, is yet a model of logical elegance and rigor : the Hke can hardly be said of the modern treatment.* * The usual algebraical treatment of proportion is not really sound. O. Henrici, Enc, Brit,, Vol. X., Geometry, § 47. MAXIMA AND MINIMA. 261 The doctrine developed in this text is purely geometri- cal, implying no numerical knowledge or calculus. It is grounded in the notions of parallelism and similarity, to stand or fall with them. Hence it will be found to have no place in bi-dimensional spherics, the doctrine of the sphere- surface, which is in many particulars quite analogous to Planimetry, the doctrine of the plane, but in which there are no similar figures. MAXIMA AND MINIMA. 344. Already, in Art. 135, the notions of maximum and minimum have been defined, but it is well to add here that not absolute but merely relative size is referred to, inasmuch as a varying magnitude may pass through a number of maxima and minima, and of these some maximum may be less than some minimum. Thus, a boy may inflate his elastic balloon till the diameter becomes 9 inches, then let it shrink to a diameter of 7 inches, again inflate it to a diameter of 12 inches, let it shrink to one of 10 inches, again inflate it, and so on. Here 9 and 12 are maxima, while 7 and 10 are minima of the diameter. Plainly, maxima and minima alternate with each other, and the course of the 262 GEOMETRY. variable may be depicted by a waving line, the values of the variable being the vertical distances of the points of the line from a fixed base-line. OF is the axis of the variable (Fig. 196); OT is the time-axis. What are the maxima and minima? How do the tangents lie at these points of the curve? 345. The general doctrine of maxima and minima calls for a method that shall seize upon the magnitude in the process of change, and subject its momentary variations to investigation. Such a method is supphed in the Infinitesimal Calculus. But there are many interesting and important geometric problems that yield even more readily and com- pletely to elementary than to more refined methods, and some of these we shall now consider. 346. What is the maximum parallelogram with given sides ? The student may easily show it to be a rectangle. Corollary. The maximum triangle with two given sides is right-angled between those sides. 347. What is the maximum triangle with given base and given vertical angle ? From the figure it is at once seen to Fig. 197. MAXIMA AND MINIMA. 263 h^ symmetric (Fig. 197). Trace the variation in both area and perimeter of the A. 348. What is the maximum triangle with given base and given perimeter ? We are sure that the symmetric A is either maximum or minimum ; for as the vertex sHps either rightward or leftward from the symmetric position by the same infinitesimal amount (Fig. 198), the two resulting A are congruent. Hence the Fig. 198. symmetric A, which lies between the two, is either greater or less than either. That it is greater, and hence is the maxi- mum, is readily proved thus : Through the vertex V draw a parallel to the base. Then no other position of the vertex can be on this parallel, as at U) for AU^ UB > AV-^ VB, as is seen at once on taking the point B' symmetric with B as to the axis VU. Still less can the vertex take a position above the parallel, as at W. Hence it must in every other position be below the parallel, as at F' ; 2in^AVB>A V'B (why ?). 264 GEOMETRY. [Th. A. 349. Theorem A (Lemma). — If the base and perwiefer of a rectili7iear figure be given, the area may be continually increased by increasing the number of sides (besides the base), and keeping them mutually equal. Data : b the given base, s the sum of the other sides. Proof. On b complete with s 3, symmetric A, a maximum. On either side take a point D distant one-third of the side AC from the vertex C. Now holding the new base BD fixed, convert BCD into a maximum (symmetric) A, keep- ing it isoperitnetric, that is, of equal perimeter. The result- ing quadrangle ADEB is greater than the A ACB (Fig. 199), and has its three sides AD, DE, EB equal. Now drawing AE, we may proceed similarly with the A ADE ; the resulting 5 -side will be greater than the 4-side, but its sides will be unequal, and we can still further enlarge the figure, keeping it isoperimetric, by equalizing the four sides. Thus we may proceed continually, and at every step enlarge the bounded area, first increasing the number of sides, and then equahzing them. As long as any two consecutive sides are unequal, we can join their ends and enlarge their A by making them equal, q. e. d. 350. Theorem B (Lemma). — Any n-side, the base being fixed and the other sides mutually equal, has maximum area only when the equal sides enclose equal angles. MAXIMA AND MINIMA. 265 Data: We consider first a 4-side, AB its base, AC, CD, DB its three equal sides, and the angles C and D equal (Fig. 200). Fig. 200. Proof. Suppose the sides to be rigid rods hinged at the angles, forming a linkage. Precisely, as in Art. 349, we know that the area is either a maximum or a minimum, and we easily prove it to be the maximum thus : Deform the linkage by thrusting the hinge C down to O ; then if AB > AC, as C descends to C on the arc of a circle, D will rise to Z>' on the arc of an equal circle. 1. Then t/ie arc CO > arc DD^. For CZ>= CD\ and CZ>' being oblique, its horizontal projection is < CD', hence the lateral or horizontal thrust of C is > the lateral thrust of D ; that is, the horizontal projection of the arc or chord CC is > the horizontal projection of the arc or chord I)D\ But even if DD' were only equal to CC, its hori- zontal projection would be greater than that of CC (why?) ; hence DD must be less than CC. Corollary. Hence AACC>BDD'. 2. The A ICC > IDDK For if the fixed length CD had slipped with its ends along the tangents at C and D into the 266 GEOMETRY. •position C"/?", then we should have had A/'CC" > VDD'^ (why?) ; much more, when the ends shp along the arcs (or chords), the end C falls lower (to C), the end D rises not so high (to Z>'), and the intersection slips further rightward (to /), the large subtractive A is increased to ICC, the small additive A is decreased to IDD\ 3. Hence the two decrements ACC and ICC produced by this deformation being respectively greater than the two increments BDD^ and IDD\ it follows that the resultant area ACD^B is less than the original area ACDB. The reasoning is not changed if we thrust in D instead of C, and it applies whatever the amount of the thrust. Hence the anti-parallelogram A CDB is the maximum. If, however, AB ' (that is, if we grant Axiom A but reject Axiom B), then we have Hyperbolic space. D. Lastly, if we grant both Axiom A and Axiom B, then we have ordinary Parabolic space. CONCLUDING NOTE. 11\ 359. The names Elliptic, Hyperbolic, Parabolic have been given by Klein. They mean lacking^ exceeding, equal- ling, and refer to a certain characteristic magnitude called the Riemannian ' measure of curvature ' (Riemann'sche Kruemmungsmaass) , which in the three cases is respec- tively negative, positive, o, or less than o, greater than o, equal to o. Instead of Klein's terms we sometimes meet with Riemannian, Lobatschevskian (or Gaussian), and Euclidian, from Riemann, Lobatschevsky and Gauss, and Euclid, — mathematicians that first set forth clearly the properties of the space-forms. 360. Some of the distinguishing features of these four spaces are the following : A. I. The ray is closed and finite. 2. The sum of the angles in a plane A is > a straight angle. 3. Two rays that meet in one point meet also in a sec- ond point. B. I. The ray is closed and finite. 2. The sum of the angles in a plane A is > a straight angle. 3. Two rays meet at most in one point only. C. I. The ray is not closed, but infinite. 2. The sum of the angles in a plane A is < a straight angle. 3. Two rays meet at most in one point only. D. I. The ray is not closed, but infinite. 2. The sum of the angles in a plane A is = a straight angle. 3. Two rays meet at most in one point. 272 GEOMETRY. 361. It is curious and noteworthy that the ray in a sim- ple Riemannian plane cuts the plane through, but not in two. For, take any ray R and two points close together, P on the left and Q on the right of R ; through P and Q draw a ray meeting R at M. Then we may pass from P to Q rightward through M\ or, since the rays are closed and meet only in M, we may pass from P \.o Q leftward and not through M\ i.e. we may pass from one side of the ray R to the other without crossing it. This may be hard for us to imagine, but perhaps not harder than for the ancients to imagine antipodes. Think of a hollow ring — a circle running all round it or across it would cut it through, but not in two. The Riemannian plane is not such a ring, but is ring-hke in being thus doubly compendent (Art. 162). 362. It will be well for the student to observe carefully just where the proof of Theorem XXXI. breaks down on rejecting Axiom B. We may then still draw through C a non-intersector of AB, making the alternates a and a' equal ; and we may also draw through C a non-intersector of AB, making the alternates fi and ^' equal. But in the absence of Axiom B, we cannot know that these two non-inter- sectors are the same ; hence we cannot know that the sum a' + 7 + )8' = a straight angle ; hence we cannot affirm that the sum a -f y + i^ = a straight angle. In fact, if the two non-intersectors are not the same, then manifestly '-i + 7 + i3 = «' + y + ^' < a straight angle. 363. Actual measurement, direct and indirect, of the angles of a A yields a sum always very near to a straight angle. But even the largest A we can construct and com- pute in the heavens are yet extremely small relatively to the whole of space, whether space be finite or infinite ; and since the defect under a straight angle, or the excess over a straight CONCLUDING NOTE. 273 angle, if there be any such defect or excess, must vary with the size of the A, in observed A it would be extremely small and so might elude our observation. In fact, for extremely small A, the only A of experience, the four spaces are so nearly alike in properties as to be indistinguishable ; just as if the earth's radius were a decillion times as great as it is, and our experience extended over no more than its present surface, we should be unable to say whether it was flat, or sphere-shaped, or egg-shaped, or ring-shaped, or saddle- shaped. 364. It appears then that the natural question. Which of the four possible homoeoidal spaces is our actual space ? is at present unanswerable. Our experience is still too narrow to enable us to decide or even to conjecture. Why, then, do we seem to prefer parabolic space, and build up our geometries on Euclid's foundations ? Because it is easier, more convenient. The superior simplicity of the Euclidian geometry is conspicuous in its doctrine of the parallel, the unique intersector, and of the sum of the angles in the plane A, which is a constant, the straight angle. The ground of our preference, then, is not a logical, but an economical one. 365. Lastly, let the student never forget that the question as to the fundamental properties of our space is at bottom a question as to the constitution of our own minds. It is they that at every instant project images of their own states and of all beings as related to them, and build up these pro- jections into the world of phenomena about us, which we call space and its contents. Space, then, is made the way our spirits make it, and to know its fundamental properties is to know fundamentally the mode in which the spirit objectifies to itself, makes an object of its own contempla- tion, the world of Not-self about it. Whence it appears that 274 GEOMETRY. not only all physical problems, but also all geometrical problems, root finally in Metaphysics. 366. In the writer's judgment the doctrine of non- euclidian spaces and of hyper-spaces in general possesses the highest intellectual interest, and it requires a far-sighted man to foretell that it can never have any practical importance. The student who would pursue the subject should read Halsted's excellent translations of Lobatschevsky and Bolyai, the Lectures and Addresses of Clifford and Helmholtz, Ball's article on Measurement in the Encyclopaedia Britan- nica, and afterwards the monographs of Riemann, Klein, Newcomb, Beltrami, Killing, and should also consult the bibliography of the subject as given by Halsted in the American Journal of Mathematics y Vols. I. and II. EXERCISES V. 1 . Central rays of a parallelogram bisect it, and conversely. 2. Central rays of any closed centrally symmetric figure bisect it, and conversely. 3. Tracts bisecting the mid-parallel of a trapezoid and ending in its parallel sides bisect it, and conversely. 4. The sums of the opposite A, into which tracts from a point within a parallelogram to its vertices divide it, are equal. 5. The sums of the opposite A formed by tracts from any point of the mid-parallel to the vertices of a trapezoid are equal. 6. Tracts from a vertex of a regular 6-side to the other vertices and the mid-points of the remotest sides divide it into six equal A. 7. The A whose vertices are the ends of one obHque EXERCISES V. 275 side of a trapezoid and the mid-point of the other is half the trapezoid. 8. The A of the medials of a A has f of the area of the A. 9. /'is a point, ABCD a parallelogram ; then AAPC = A AFD ± A AFB. 10. The joins of the mid-points of the opposite sides of a 4-side concur with the join of the mid-points of the diagonals. 11. * Divide a A or parallelogram into two parts in the ratio /:;//. 12. Divide a parallelogram by tracts from a vertex into n equal parts. 13. Divide a A into n equal parts by tracts from a point on a side. 14. Transform a A or parallelogram into another of same base with a given angle at the base. 15. Transform a A or parallelogram into another with given base. 16. Transform a A or parallelogram into another with given base and given adjacent angle. 17. Transform a trapezoid into an anti-parallelogram of the same altitude. 18. The common border of two areas lying between two rays is a broken line ; rectify this border without affecting the areas. 19. Inscribe in a given circle a rectangle of given area. 20. If tracts between the parallel sides of a trapezoid, not intersecting within it, divide the mid-parallel into n equal parts, they also divide the area into n equal parts. 21. In a trapezoid, the join of the mid-points of the diagonals equals the difference of the parallel sides. 22. Construct two tracts, knowing their ratio and their sum or difference. 23. Investigate the proportionaHties between the seg- 276 GEOMETRY. ments of the altitudes of a A made by the orthocentre and the segments of the sides made by the altitudes. 24. To the base of a A draw a parallel cutting off a A of given perimeter. 25. * The sum of two similar figures on the sides of a right A equals a third similar figure on the hypotenuse. 26. If intersecting semicircles be drawn on all the sides of a right A, the sum of the two small crescents will equal the larger (Hippocrates, 450 b. c). 27. Bisect a A by a parallel to its base, and generalize the problem. 28. The rectangle of the segments into which one alti- tude of a A is cut by the others is the same for all the altitudes. 29. Construct a A, knowing its altitudes. 30. The medials of a A cut it into six equal A. 31. A's mid-rays cut sides into irjslt ; 1tJ=iJ-\-jl-\- li= ^rs. 32. The altitudes of a A cut it into six A, so that the sums of the alternate A are equal. 33. Through the perimeters of an inscribed and a circum- scribed regular /z-side express those of the 2 ;z-side. 34. The area of a ring between two concentric circles equals that of a circle having as diameter the chord of the outer circle tangent to the inner circle. 35. The radius of the earth being 6370 kilometers, how far might one descry a ship from the summit of Chimborazo (21,424 feet) ? 36. How many hills of corn one yard apart may be planted in a rectilinear field of one acre ? 37. Chords from any point of a circle to the ends of a diameter divide its conjugate chords harmonically. 38. Chords from any point of a circle to the ends of a chord divide its conjugate diameter harmonically. EXERCISES V. 277 39. Transform a A into another A' similar to a given A". 40. Find the locus of a point whose distances from two given rays are in a fixed ratio. 41. All rays whose distances from two fixed points are in a fixed ratio envelop two fixed points ; find them. 42. On a given ray find a point whose distances from two fixed points (or rays) are in a fixed ratio. 43. Through a given point draw a ray whose distances from two fixed points shall be in a fixed ratio. 44. Find a point (or points) whose distances from three fixed non-concurrent rays L, M, iV shall be as /: w : 71. 45. Find a ray (or rays) whose distances from three non- collinear points A, B, C shall he 2iS a-, b -.c. 46. Find a point whose distances from three fixed points shall he 2iS I : m '. n. 47. In every trapezoid the mid-points of the bases along with the intersections of the non-parallel sides and of the diagonals form an harmonic range. 48. The locus of a point whose distances from A and B are in the ratio ^ : ^ is a circle on AB as central ray and dividing AB harmonically in the ratio a : b. 49. The locus of a point whence the collinear tracts AB and BC appear equal is a circle dividing AC harmonically. 50. Find the point whence three collinear tracts AB, BC, and CD appear equal. 51. /'is a given point in a given angle BAC \ draw BC so ihdXPB.BC'. :l\m, or BP.PCwl.m, or BA.AC:: l\ m. 52. From the continued ratio of the sides determine the continued ratio of the altitudes of a A. 53. A marble rests in a conical glass ; another rests on it and touches the glass all around ; and another, and so on. How are the radii of the marbles related in size ? 278 GEOMETRY. 54. Find the area bounded by three equal tangent circles. 55. Compare the A of the centres with the A of the com- mon tangents of three tangent circles. 56. In a regular A is inscribed a circle ; tangent to it and two sides of the A, another circle ; and so on. How do the radii decrease? 57. J/ is the mid-point and P any other point of the tract AB ; semicircles are drawn on AM, MP, AP, and PB, all on the same side of AB ; show that the sum of the first and second equals the fourth plus the area bounded by the first three. 58. Inscribe a circle in a given sector of a circle. 59. Find a point on a circle whose distances from two chords are proportional to the chords. 60. The distance of any point of a circle from the chord of contact of two tangents is a mean proportional between its distances from the tangents. 61. When does the altitude to the hypotenuse of a right A divide the hypotenuse in extreme and mean ratio ? 62. In a regular 5-side each diagonal is divided in extreme and mean ratio by two others and all form a regular 5-side. Compare the areas of the two 5 -sides. 63. The rectangle of the distances of any point of a circle from two opposite sides of an encyclic 4-side equals the rectangle of the distances from the diagonals. 64. RA and RB are two equal rigid rods pivoted at R ; 6>(2 is a rigid rod pivoted 2ii O dJid 0Q= OR ; AQBP is a rhombus formed of equal rigid rods movable about their junction-points. Show that, as Q traces a circle about O, P traces a ray normal to the ray OR, and that P and Q are inverse as to R. N.B. Such is the mechanical invertor called Peaucellier's ceU. EXERCISES V. 279 65. In any range of four points, as ABCPj show that AB' CF+BC'AF^ CA- BP=o. 66. In any pencil of four tracts, as OA, OB, OC, OD, show that AAOB-ACOD-\-ABOC'AAOD-^ACOA'ABOD=o. 67. The diagonals of a parallelogram concur with the diagonals of its complemental parallelograms. 68. The mid-points of the diagonals of a 4-side are collinear. 69. Rays through the vertices of a A and the points of touch of the ex-circles concur. 70. Normals to the sides of a A, at the points of touch of the ex-circles, concur. 71. When normals from the vertices of one A to the sides of another A' concur, so do normals from the vertices of A' to the sides of A. 72. Normals to the three sides of a A through the points of touch of two ex-circles and the in-circle concur. 73. The feet of normals from any point of a circle to the sides of an inscribed A are collinear (on Simson's Line). 74. Rays through the vertices of a A and the points of touch of the in-circle concur. 75. Mid-rays of the angles of a A (three outer, or two inner and one outer) intersect the opposite sides colHnearly. 76. Tangents to the circumcircle of a A at its vertices intersect the opposite sides collinearly. 77. If the joins of the vertices of a A with three inter- sections of the opposite sides by a circle concur, so do the joins of the vertices with the other three intersections of the opposite sides. 78. Find a circle as to which two pairs of collinear points are inverse. {Hi?tt Draw a circle through each pair, and also their power-axis.) When is the circle sought real? 280 GEOMETRY. 79. Find the inverse of a given point as to a given circle. 80. Given two points, find a circle of inversion having given radius or given centre. 81. A circle through a pair of inverse points cuts the circle of inversion orthogonally, and conversely. 82. The squared distances of a point on the circle of inversion from two inverse points are in the ratio of the central distances of the points. Z-^. The power-axis of a fixed circle and a variable circle through two inverse points as to the fixed circle envelops a fixed point. 84. Find the inverse of a circle when it passes through and also when it does not pass through the centre of inversion. 85. When does a circle invert into itself ? 86. A circle, its inverse, and the circle of inversion have a common power-axis. 87. Inversion does not change the size of the angle under which two lines intersect. ZZ. The 9-point circle of a A touches the in- and ex-circles of the A. 89. The joins of the intersections of two circles with their common diameter and a common orthogonal circle concur on that orthogonal. 90. The join of the polars of two points is the pole of the join of the points. 91. If a pole trace the sides of an ;7-side, P, the polar will envelop the vertices of an //-angle, Q\ and if the pole trace the sides of Q, the polar will envelop the vertices oiP. 92. If the pole trace any line Z, the polar will envelop a corresponding line V ; and if the pole trace Z', the polar will envelop Z. EXERCISES V. 281 Defs. Two lines, either of which is enveloped by the polar, as the other is traced by the pole, are called reciprocal. When the reciprocals coincide, the figure is called self- reciprocal or self-conjugate, while the centre and the circle of reference are called the polar centre and the polar circle of the figure. 93. If a A has a polar centre, it is the orthocentre. 94. If the polar circle is real, the A is obtuse-angled. 95. The polar circle of a A is orthogonal to the circles on the sides as diameters. 96. Invert the sides of a A as to its polar circle. 97. The ends of a diameter of a circle are conjugate as to every orthogonal circle. 98. As the orthogonal circle (of 97) varies, the polar of either diameter-end envelops the other. 99. The distances of any two points from a polar centre vary as the distances of each from the polar of the other as to that centre (Salmon) . 100. Polar reciprocal A are in perspective. Def. Circles, every pair of which have the same power- axis, are called co-axal. loi. If two circles intersect in A and B, all co-axals go through A and B. 102. The contrapositive of loi. Hence there are two kinds of co-axals : common point co-axals (loi) and non-intersectors, so-called limiting point co-axals (102). 103. The centre line of common point co-axals is a ray, namely, the common power-axis of co-axals orthogonal to the common point co-axals. 104. One and only one of these orthogonals goes through every point of the plane except the common points, which are therefore called (Poncelet) limiting points of the orthogonal co-axals. 282 GEOMETRY. 105. Draw a double system of mutually orthogonal co- axals. 106. The polars of a point as to co-axals concur, and conversely. 107. The difference of the powers of a point as to two circles is proportional to the distance of the point from the power-axis. 108. The locus of a point whose tangent lengths from two circles are in fixed ratio is a co-axal circle. 109. When does the power-centre of three circles become indefinite, and when does the radius become imaginary? 1 10. The polar centre of a A is the power-centre of three circles on any tracts from the three vertices to the sides as diameters. 111. Three collinear points on the sides of a A being joined with the vertices, the circles on these joins as diameters are co-axal. 112. The four polar centres of the four A, formed by the sides of a 4-side taken in threes, are collinear on the power- axis of circles on the diagonals of the 4-side as diameters. 113. Invert a double orthogonal system of co-axals; as special case take a common or Umiting point as centre. 114. Show that any two circles may be inverted into equal circles, and find the locus of the centre of inversion. 115. Invert three non-co-axal circles into three equal circles. 116. Draw a circle touching these three equal circles, and re-invert the four circles. What problem is hereby solved ? 117. Discuss the various possible positions of the centres of similitude of two circles. Def. The circle of similitude of two circles has the tract between the centres of similitude as diameter. 118. The circle of similitude is co-axal with the two circles. EXERCISES V. 283 119. From any point on the circle of similitude the two circles appear to be of the same size. 120. Find a circle as to which the power-axis and the circle of simiUtude invert into each other. Defs. A tract with its ends on a figure is called a chord of the figure. A ray bisecting a system of parallel chords is called a diameter of the figure. Two diameters, each halv- ing all chords parallel to the other, are called conjugate. 121. Every two-ray (^Zweistrahl), or angle considered not as a magnitude but as a figure, has an infinity of diameters, namely, every ray through its vertex. 122. A A has three diameters, namely, its medials. 123. A parallelogram has two pairs of conjugate diameters. 124. A two-ray has an infinity of conjugate diameters. 125. If Z' and ^' be conjugate diameters of (the two- ray) LM, then L and M are conjugate diameters of (the two-ray) L'M'. Def. Four such rays are called harmonic, because : 126. They cut every transversal in four harmonic points. 127. Conversely, four concurrent rays through four col- linear harmonic points are harmonic. 128. The join of an outer vertex with the intersection of the inner diagonals of a 4-side cuts two opposite sides, each in a fourth harmonic to the three vertices on the side. Hint. In Fig. 54, let / be the intersection of the inner diagonals CE, DF; let a 4th harmonic through A cn\. BC and BD at H and K. Then IV, IE, IB, IK are four har- monic rays, and so are IF, IC, IB, IB; also ID, IE, IB, are the same rays as IF, IC, IB ; hence IH and IK are the same ray ; hence the 4th harmonic through A goes through /. 129. Enumerate the harmonic ranges and pencils in 128. 130. A system of co-axals determines the points of every ray in pairs of conjugates, F and F', Q and Q, so that 284 GEOMETRY. IP • IP' = IQ • IQ', where / is the intersection of the ray with the power-axis. Defs. Points so determined are said to form an Involution. The fixed point / is called the centre of the Involution, the constant product or rectangle of the central distance of the conjugates is called the power of the Involution, and is posi- tive or negative according as the distances are like-sensed or unlike-sensed. 131. An Involution is determined by its centre and a pair of conjugates, or by two pairs of conjugates. 132. When the power is positive there are two self-conju- gate points (called double points or foci), and the focal tract is divided harmonically by every pair of conjugates. 133. Conversely y all pairs of points dividing a tract FF^ harmonically form an Involution with F and F as foci and the mid-point / of FF^ as centre. 134. When the power is negative there are no (real) foci but there are two conjugate points, E and E\ equidistant from the centre. Def. Rays of a pencil passing through an Involution of points form an Involution of Rays. 135. Develop and express the reciprocal properties cor- responding to 1 3 1-4. 136. In ge?ieral, the points of a row and the intersections of their polars with the axis of the row form an Involution. Hint. P the point, L the axis, O the centre of the referee circle S, which cuts L at i^and F\ Q the intersection of L with the polar of P, 01= d= distance of L from O, OA = r= radius of 6* on OF. Draw a circle K on FQ as diameter about C cutting OF at R. Then, difference of powers of O and / as to A" is OC' -JC'=~0~f = r'-IQ'IF. Hence r- — d'^ = IQ • IF= a constant, q. e. d. EXERCISES V. 285 137. What exception does 136 suffer? What are the relations of the three possible cases ? 138. Two polar conjugate points (or rays) and the pole (or polar) of their join determine a polar reciprocal A. 139. Conjugate points on a ray (or rays through a point) form an Involution. When positive? When negative ? 140. Two conjugate points in a secant divide the chord, and two conjugate rays through a point divide the angle of tangents from the point, harmonically. 141. The outer vertices and the intersection of inner diagonals of an encyclic 4-side form a polar A. 142. The diagonals of a pericyclic 4-side form a polar A. 143. Employ 141 and 142 to find the polar of a given pole and the pole of a given polar by use of ruler alone. 144. Given a centre of similitude of two circles, find its polars as to the circles and the power-axis by use of ruler alone. 145. Given the power-axis of two circles, find its poles as to the circles and the centres of similitude by use of ruler alone. Defs. Two figures are said to be in perspective where the joins of corresponding points all go through a point called the centre of projection. — The rays are called rays of pro- jection. Parallel rays are thought concurrent at 00. — Two pencils are said to be in perspective when the joins of cor- responding rays all lie on a ray, called the axis of projection. — The centre of projection and the axis of projection are plainly self-correspondent. — Each system of points is also said to be in perspective with the pencil of rays through them. — In elementary work we impose the condition that col- linear points in the one figure shall correspond to collinear points in the other. 286 GEOMETRY. 146. Two tracts are always in perspective as to two cen- tres. 147. Express and prove the reciprocal theorem as to angles (two-rays). 148. Three collinear points and three concurrent rays are always projective, i.e., may always be brought mio perspective. 149. Two triplets of colHnear points are always in pro- jection. (For it is enough to slip the triplets each along its ray till a pair of correspondents coincide.) 150. State and prove the reciprocal theorem for pencils of three. Def. The ratio of the distances of any third ray iV of a pencil from two base-rays L and M of the pencil is called the distance-ratio of the third ray as to the other two ; it may be written ( ) and is reckoned + or — according \NM) as the angles LN and NM are reckoned in the same or in opposite wise. 151. Trace the course of the distance-ratio as the third ray completes a rotation about the centre of the pencil. 152. The distance-ratio equals the sine-ratio of the angles formed by the third ray with the base-rays. 153. When two distance- or sine-ratios are counter, the four rays are harmonic, and the ratio of their ratios is — i. 154. Compare distance-ratios of corresponding angles and tracts in a pencil. Def. The ratio of two distance-ratios in the same pencil, whether of tracts or of angles, is called the cross-ratio, or ratio of double section (^Doppelschnittsverhaeltniss), or an- harmonic ratio, of the bounding points or rays, and is written {A BCD) or {LMNF). Its value is ABj_CD sin LM- sin NP BC'DA sinMN'Sm^ EXERCISES V. 287 155. The joins of the vertices of a A with three colHnear points on the opposite sides divide the angles so that the triple sine-ratio = — i . 156. Three concurrent rays through the vertices of a A divide the opposite sides so that the triple distance-ratio = 4- I {^Ceva, 1678). 157. Convert these two theorems and those of Arts. 158. Apply these converses in establishing the concur- rences of altitudes, mid-normals, mid-rays, medials, etc. 159. The sides of a A cut by a circle in six points are divided so that the continued product of the distance-ratios is -f I {Carnot, 1 753-1823). 160. Express and prove the corresponding proposition concerning six tangents, drawn from three points, to a circle {ChasleSf 1850). Nint. r the radius; A, B, C the points; AT^, AT^ two tangents-lengths = /j, 4 ; ^u ^2 the distances of T^, T^ from BC ; /i, I2 the intersections of two tangents, parallel to BC, with the rays AT^, AT2', d, e, f the distances of BC, CA, AB from the centre O ; jPthe projection of o on BC. Then a^: ti=^ d — r\ AIi, a^,'. ti=- d -\- r \ AI^, :, aya.2 : A4 = d^ — r^ : AI^ • AI^. But from similar A OAIi, OAL we have AI- AI^ = ACf ; :. a^a^ \ tit^—d'—r^ \ A O^. Similarly, byd2 : tit2=^—r- : AO'. .'. a^ai : ^A — d'^ — r'^ : e^ — r^. Find two analogous equa- tions, combine the three by multiplication, and the propo- sition in question results. 161. The three joins of the opposite sides of an encyclic 6-side are collinear {Pascal, 1640). Hint. Let. i 2 3 4 5 6 be the 6-side ; I, J, K the joins 288 GEOMETRY. of the opposite pairs, 12 and 45, 23 and 56, 34 and 61. The alternate sides 61, 23, 45 form a A ABC, and are cut collinearly by the other alternate sides 12, 34, 56; apply thrice the theorem of Menelaos; multiply, cancel, and apply the converse of the theorem of Menelaos. 162. Express and prove the corresponding theorem of Brianchon (1806), using Ceva's theorem and converse. 163. Every different order of sides respecting vertices gives a different hexagram of Pascal respectively hexagon of Brianchon ; how many of each are possible ? 164. These so-called Pascal Y2,y^ are concurrent, and the Brianchon points are colhnear, in sets of three {Steiner, 1832). 165. Apply the theorems of Pascal and Brianchon to find with ruler alone a tangent to a given circle at a given point and the point of touch of a given tangent {Steiner, 1833). Def. In projecting one ray L on another Z', there will be one ray of projection/^r«//(?/to L' and meeting L at V. To this point V, and to it only, there corresponds no finite point of Z' ; to points close at will to V there correspond points far at will on Z'. Hence V is called the vanishing point of Z with respect to Z'. Similarly, £/' is the vanishing point of Z' as to Z. 166. /* and Z^ correspond on Z andZ'; show and state in words that FV - F'U'= OF- OU'. Def. This constant rectangle (product) is called the constant of projection. 167. Two A are always projective. (For we can always place a pair of vertices on a point, or a pair of sides on a ray, and — what then ?) 168. Three pairs of points taken at random on three con- current rays, each pair on a ray, determine two perspective A whose corresponding sides meet collinearly. (Use the propositions of Menelaos and Ceva.) EXERCISES V. 289 169. Express and prove the reciprocal of 168. 1 70. The Locus of the vanishing points of all rays of one of two (rectilinearly) perspective figures is a ray (called vanishing ray) parallel to the Axis of projection. 171. The distance of the one vanishing ray from the Axis equals the distance of the other from the Centre. 172. Parallel rays of one of two perspective figures corre- spond to rays concurrent in a vanishing point of the other. 173. A circle is in perspective with itself, any pole and corresponding polar being Centre and Axis. 1 74. The vanishing ray halves the distance between the Centre and the Axis and is the Power-axis of the circle and the Centre of projection (regarded as a point-circle). 175. Two circles are in perspective as to a centre of similitude, and the mid-parallel of the polars of this centre is the Axis. 1 76. A figure F is pushed and turned about in a plane into any other position P \ show that the same change of position may be effected by simply turning about a point in the plane called the Centre of Rotation. Hint. A, B, C three points of F, and A\ B\ C their positions in F ; draw the mid-normals ofAA', BB\ CO ; etc. 177. If the joins of the corresponding vertices of two A be concurrent, the joins of the corresponding sides are collinear; and conversely (Desargues). 1 78. If a quadrilateral be inscribed in a circle Q while two of its opposite sides touch a circle C^ and the other two touch a circle C3, then the three circles Cj, C^, C^ are co- axal, — a theorem very important in the theory of Elliptic Functions. 290 GEOMETRY. 179. The area of a pericyclic polygon equals half the rectangle of the in-radius and the perimeter. 180. If i- be the half-sum of the sides a, b, c oi 2i A, and ^1 ^15 ^2> ^3 the in- and ex-radii, then A = rs = ri{^s — a) = r,{s-b)=r^{s-c). 181. Hence, show that - = - H 1 — and A^ = rr^r^r^,. r ri r^ r^ 182. \i a,b,c\yt the sides of a A, and h the altitude CO, show from a- — F^ -\- r — 2 c - AC that /^2 = 54^V - (^2-f c'~a^y\l4,c' = ^^s(s - a) {s- b) {s-c), whence A^ z=s{s — a) {s — b) {s — c) {Hero, 250 B.C.) . 1 83. If a, b, c be the sides of a A, and m the tract from C to ^ halving ^ C and cutting c into parts u and v, show that ab — Mv= — ^ „ - sis — c^. 1 84. If two such tracts in a A be equal, the A is sym- metric. 1 8s. Show that the circum-radius of a A = . 4A 186. Express through r the radius of a circle, the sides and areas of the regular inscribed and circumscribed 6-sides, 4-sides, 3-sides, lo-sides, 5 -sides; also the apothegms of the in-polygons. 187. Given the centre of similitude and two correspond- ing rays of two similar figures in perspective, find P^ corre- sponding to a given P. 188. Corresponding angles of similar figures in perspec- tive have always the same sense. 189. If the A ABC, ABD, etc., of F are similar to AB'C, A'B'D\ etc., of F\ then 7^ and F' are similar. 190. Two similar figures are in perspective when two corresponding rays are parallel. EXERCISES V. 291 191. Through any point /* (or -parallel to any ray iV), draw a ray towards the inaccessible intersection of the rays L and M. 192. The sides of a quadrangle are given in position; draw the diagonals when two opposite vertices are inacces- sible, and when all the vertices are inaccessible. 193. Draw a circle S^ in perspective with S as to the centre O, so that two given points /'and /" shall correspond ; so that two given parallels L and V shall correspond ; so that S shall have a given radius r' ; or, so that the centre of S shall lie on a given ray. 194. Draw S tangent to .S so that two given points P and P, or two given rays L and L\ may correspond. 195. Draw a circle to touch a given circle and also touch a given ray at a given point. 196. Draw a circle tangent to two given rays and a given circle. 197. In any pencil of four rays, as OA, OB, OC, OP — written 0{ABCP)—AOB\ - COP\^- BOC]- AOP\ + COA\'BOP\ = o. Bint. Note that AB 'p= OA- OB • AOB\, etc., and use 66. 198. The concurrent rays L, M, N are distant /, m, n from P; prove /• MN\ -\- 7n - NL\ -\- n - LM\ = o. 199. If 2s = a-\-b-\-c-\-d= perimeter of an encyclic quadrangle, show that A^ = (^ — a) {s — b){s — c) {s — d) and express this result symmetrically through a, b, c, d. 200. If r and r' be the circum-radius and in-radius and 2s = a -{-b -\- c the sum of the sides of a A, prove that 2 rr^s = abc. 201. Draw a fourth harmonic to three rays of a pencil. 292 GEOMETRY, 202. The sum of the distances of the sides of a A from a point, each side multiplied by the sine of its opposite angle, is constant for that A. 203. A ray cuts the sides a, b, c oi 2, /\ under angles a , P', y' ; show that a\' a'\ -{-^l- 13'\-\- y\'y'\ = o. 204. Concurrent rays through the vertices of a A divide the sides so that the continued product of the ratios of division is — i. 205. Concurrent rays through the vertices A, B, C of a, A cut the sides at P, Q, R ; show that the intersections /, /, K of AB and FQ, BC and QR, CA and RP are collinear; also that' if BJ and CK, CK and AI, A I and ^/meet in S, T, U, then AS, BT, C^ concur. INDEX. (The numerals refer to pages. Only the more important references are given.) Abstraction, 7. Addends, 17. Addition, 17. theorem, 248. Adjacent, 30. Alternation, 172. Alticentre, 68. Altitude, 67, 148. Ambiguous case, 46. Angle, 19, 72, adjacent, 30. alternate, 54, complemental, 32. corresponding, 53. explemental, 31. interadjacent, 54. right, 31. round, 26. straight or flat, 26. supplemental, 31. vertical, 37. Anomaly, 191. Antecedents, 172. Anti-homologous, 217. Anti-parallelogram, 62, 65. Apollonius, 225. Apothem, 249. Arc, 13, 90. Area, 144. Areal unit, 231. Arms, 19. Axal symmetry, 79. Axally symmetric, 77. Axis, power or radical, 213. of projection, 285. of similitude, 216. of symmetry, jj. Babylonian, 70, 133. Band, 86. Base, 37, 148. Bases, major and minor, 65. Beltrmni, 2j2. Bi-dimensional, 5. Bisect, 125. Bisector, 33. Bolyai, 269, 270. Border, 5. Boundary, 252. Boundless, 2. Brianchon, 288. Carnot, 287. Cell, Peaucellier^s, 278. Central symmetry, yj. Centre, 18. ■ of circle, 11, 93. of inversion, 168. of involution, 284. of pencil, 80. power-, 168, 213. of projection, 285. of symmetry, 77. 293 294 GEOMETRY. Centric figure, 215. Centroid, 66. Ceva, 287, etc. Chasles, 287. Chord, 90, 283. of contact, 108. Circle, 13, 90. of inversion, 168. of similitude, 282. Circum-circle and centre, 67, 96. Clifford, 274. Clockwise, counter-clockwise, 25, 71, 72. Closed, 72. Co-axal, 281. Collinear, 82. Common point, 281. Commutative, 17. Compasses, 96, 192. Compendent, 145. Complanar, 25. Complement, -al, 32, 96, 155. Compounded, 172. Conclusion, 25. Concur, concurrent, 65. Congruent, 16, passim. Conjugate, 93, 94, 222, 283. Consequence, 172. Continuous, 3. Contra-perspective, 188. Contra positive, 40. Convex, 72. Corollary, 22. Correspond, -ent, 16, 76, 145, 173. Cosine, 241. Cosines, law of, 245. Couplet, 172. Criteria, 147. Critical, 107. Crossed, 63. Cross-wise, 216. Cross-ratio, 286. Cuboid, 239. Dase, 253. Definites, 234. Definition, 59. Degrees, 70. Denominator, 228. Diagonal, 57, 63. Diameter, 93, 215, 283. Difference, 17, 172. Dimensions, 3, 149. Direct perspective, 188. Dissimilarly, 219. Distance, 15, 20. ratio, 286. Divided, 172. , similarly, 179. Division, harmonic, 184. , inner and outer, 182. Double, 78. Eidograph, 192. Ellipse, 94. Elliptic, 270, 271. Encyclic, 64, passim. Enthymeme, 30. Envelope, 119. Equal, -ity, 19, 147. Equator, 6. Equiangular, 70. , mutually, 173. Equiareal, 267. Equidistant, 11. Equilateral, 70. Equivalent, 49. Erotetic, 27. Euclid, -ian, 55, 159, 255, 271. Euler, 141. Even, 242. Ex-centre and circle, 69, 173. Explemental, 30, 95. Extreme and mean ratio, 202. Extremes, 171. Family, 132. INDEX. 295 Fermat, 225. Feuerbach, 112. Figure, 16. Flat angle, 26. Foci, 284. Four-side, 63. Fraction, 228, Frischauf, 270, 274. Function, 240, 242. Gaultier, 168. Gauss, 204, 271. Generated, 210. Geodetic, 269. Geometric mean, 172. Gergonne, 225. Golden section, 202. Half-strip, 87. Halsted, 269. Hankel, 234. Harmonic, 183, 2Z2, passim. Helmholtz, 274. Henrici, 260. Heptagon, 116. Hero, 290. Hexagon, hexagram, 288. Hippocrates, 276. Homoeoidal, -ity, 2, passim. nomothetic, 188. Hyperbola, 94. Hyperbolic, 270, 271. Hyper-euclidean, 55. Hyper-spaces, 274. Hypotenuse, 68. In-centreand circle, 68. Incommensurable, 231. Infinite, 2, 252. Infinitesimal, 250. Inner, innerly, 33, 63. Inscribed, 100. Instruments, 192. Inverse, 63, 216. Inverse points, 168. Inversion, 216. Inverted, 172. Invertor, 278. Involution, 284. Irrational, 234. Isoclinal, 79. Isoperimetric, 264, 267. Isosceles, 37. Joins, 76. Kaleidoscope, 137. Killing-, 274. Kite, 83. Klein, 271. Kriimmungsmaass, 271. Lemma, 92. Length, tangent-, 108, 213. Limit, 98, 252, 267. Limiting points, aSi. Line, 6. Linkage, 265. Lobaischevsky, -an, 271. Locus, 12. Magnitudinal unit, 228. Maximum and minimum, 165, 261. Means, 171. Measure of curvature, 271. Medial, 38. Median section, 202. Menelaos, 239, 288. Metric number, 228. M-fold, 226. Mid-normal, 37. Mid-parallel, 65. Mid-ray, 33. Minutes, 70. Montyon, 121. Multiple, 226. 296 GEOMETRY. N-angle, 72. Newcomb, 274. Nine-point circle, 112. Non-intersectors, 54. 5lormal, 31, 63, 102. Not-self, 273. N-side, 72. Numerator, 228. Numerics, 234. Odd, 245. Open, 72. Operation, laws of, 234. Origin, 71. Orthocentre, 68. Orthogonal, 108. Outer, outerly, 33, 63. Pantagraph, 192. Pappus, 209. Parabolic, 270, 271. Parallel, 55. Parallelogram, 57. Pascal, 287. Peaucellier, 121, 278. Pencil, 80. Pergae, 225. Peri cyclic, 117. Perimeter, 74, 115. Perimetric ratio, 253. Period, -ic, -icity, 242. Peripheral, periphery, 99. Permanence, 234. Perspective, 188, 285. Plane, 11. Polar, pole, 108, 219, etc. Polar centre and circle, 281. Polygon, 71. Point, 6. Ponce let, 281. Porism, 22. Postulate, 23, 122. Power, 166, 213. Power-axis and centre, 168, 213. Premisses, 29. Principle, 234. Problem, 120. Product, 235. Projection, 160, 286. , constant of, 288. Proportion, etc., 16^, passim. Ptolemy, 180. Pythagoras, 159. Quadrilateral, 63. Radian, 255, Radical axis and centre, 168. Radius, 95. vector, 2IO. Ratio, 181. of double section, 286. cross, distance, sine, 286. of similitude, 213. Ray, 14. of projection, 285. Reciprocity, reciprocal, 80, 281. Rectangle, 52, 149. Reentrant, 72. Referee, 220. Regular, 70. Reversible, reversibility, ii, 79. Rhombus, 60. Rotation, centre of, 289. Riemann, -ian, 11, 145, 271. Row, 80. Salmon, 281. Secant, 90. Seconds, 70. Sect, 16. Section, ratio, 286. Sector, 95, 192. Segment, 95. Self-conjugate, 281. correspondent, 285. -reciprocal, 281. INDEX. 297 Semi-circle, 95. Seven-side, 116. Sexagesimal, 134. Sextant, 137. Shape, 56. Similar, -ity, 77, 175, 188, 213. Similitude, axis and centre of, 213, 216. Simson's line, 279. Sine, 240. Sines, law of, 245. Sine-ratio, 286. Size, 56. Small at will, 230. Solid, 8. Space, I. Spaces, four forms of, 270. Sphere, 11. Spherics, 261. Square, 61, 157. Squaring circle, 253. Steiner, 288. Strip, 86. Subtend, 90. Subtraction, 17. Sum, 17, 146, 172. Summand, 17, 146. Supplemental, 31, 96. Surface, 5. Surveying, 246. Symmetric, 77, 82. Symmetry, axal and central, 76, 77. , axis and centre of, 'jt. System, 132. Taction-problem, 212. Tangent, 102. length, 108, 213. Terms, 171. Theorem, 22. Three-side, 87. Time- axis, 262. Tract, 16. Trapezoid, 65. Triangle, 35. Triangles, similar, 56. Triply laid, 87. Two-ray, 283. Unit-magnitude, 228. Vanishing point and ray, 288, 289. Vertices, 35, 72. Vieta, 225. Westings, 246. Year, 70. Zweistrahl, 283. INTRODUCTORY MODERN GEOMETRY OF THE POINT, RAY, AND CIRCLE, BY WILLIAM B. SMITH, Ph.D., Professor of Mathematics in Missouri State University. Now Ready. Complete edition, $i.io. Copies of this complete edition will he exchanged for such copies of Part I. (75 cents) as are returned to the publishers in good condition, on payment of the difference in price. The work follows the lines struck out by the great geometers of the last half century, and presents the subject in the light of their researches. The text proper conducts the student through the taction problem of Apollonius, while the exercises direct him much further in the doctrines of perspective and projection. This book is written primarily for students preparing for admission to the freshman class of the Missouri State University, and has already been thoroughly tested in the sub-freshman department of that institution. 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Vyvyan, M.A. 3d ed., revised and corrected. 90 cts. WARD. — Trigonometry Examination Papers. 60 cents. WOLSTENHOLME. — Examples for Practice in the Use of Seven-Figure Logarithms. By Joseph Wolstenholme, D.Sc. 8vo. $1.25. i,t% Keys are sold only upon a teacher's written order. MACMILLAN & CO., 112 Fourth Avenue, New York. ELEMENTARY SYNTHETIC GEOMETRY OF THE POINT, LINE, AND CIRCLE IN THE PLANE. By Nathan F. Dupuis, M.A., F.R.C.S., Professor of Mathematics in Queen's College, Kingston, Canada. 16mo. |;1.10. FROM THE AUTHOR'S PRKFACE. " The present work is a result of the author's experience in teaching geometry to junior classes in the University for a series of years. It is not an edition of 'Euclid's Elements,' and has in fact little relation to that celebrated ancient work except in the subject-matter. "An endeavor is made to connect geometry with algebraic forms and symbols : (1) by an elementary study of the modes of representative geometric ideas in the symbols of algebra ; and (2) by determining the consequent geometric interpretation which is to be given to each inter- pretable algebraic form. ... In the earlier parts of the work Con- structive Geometry is separated from Descriptive Geometry, and short descriptions are given of the more important geometric drawing instru- ments, having special reference to the geometric principle of their actions. . . . Throughout the whole work modern terminology and modern processes have been used with the greatest freedom, regard being had in all cases to perspicuity. . . . " The whole intention in preparing the work has been to furnish the student with the kind of geometric knowledge which may enable him to take up most successfully the modern works on analytical geom- etry." " To this valuable work we previously directed special attention. The whole intention of the work is to prepare the student to take up suc- cessfully the modern works on analytical geometry. It is safe to say that a student will learn more of the science from this book in one year than he can learn from the old-fashioned translations of a certain ancient Greek treatise in two years. Every mathematical master should study this book in order to learn the logical method of present- ing the subject to beginners." — Canada Educational Journal. MACMILLAN & CO., 112 FOUETH AVENUE, NEW YORK. 10 UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. Fine schedule: 25 cents on first day overdue 50 cents on fourth day overdue One dollar on seventh day overdue. RBC'D »-0 29May62SM REC'D LD MAY 1 5 1962 MAR 1 8 196G 3 LD 21-100m-12,'46(A2012si6)4120 ^^C'D i,0 /S'33 THE UNIVERSITY OF CAUFORNIA LIBRARY