BOOK VII. MODERN GEOMETRY. CHAPTER I.' TRANSVERSALS. 522,. In modern geometry the sect AB is distinguished from the sect BA as of opposite sense, so that BA = — AB, and thus when ABC are costraight the ratio AC/BC is never the same for two positions of C. 522,. [Menelaus.] If the sides of the triangle ABC, or the sides produced, be cut by any transversal in the points a, b, c, respectively, then [Ab/bC][Ca/aB][Bc/cA] = I. α 3 D A FIG. 232. Inversely, given this relation, the points a, b, c will be costraight. Proof. Draw BD || to AC, and meeting the transversal in D: then Bc/cA = DB/Ab, and Ca/aB = bC/BD; therefore [Ca/aB][Bc/cA]=[bC/BD][DB/Ab]=-bC/Ab=−1/[Ab/bC]. Inversely, if the straight ab meet AB in c', then by Menelaus, [Ab/bC][Ca/aB][Bc'/c'A] = But by hypothesis, [Ab/bC][Ca/aB][Bc/cA] Therefore c and c' coincide. 1. 523. Corollary. If a traversal intersects the sides AB, BC, CD, etc., of any polygon in the points a, b, c, etc., in order, then [Aa/aB][Bb/bC][Cc/cD][Dd/dE] . . . etc. = 1. Proof. Divide the polygon into triangles by straights through one vertex, apply Menelaus to each triangle, and combine the results. B FIG. 233. A 524. [Ceva.] If the sides of triangle ABC are cut by AO, BO, CO in a, b, c, then [Ab/bC][Ca/aB][Bc/cA] = 1. Inversely, given this relation, the straights Aa, Bb, Cc will be concurrent. Proof. By the transversal Bb to the ▲ AaC we have [Mene laus] [Ab/bC][CB/Ba][aO/OA] and by the transversal Cc to the ▲ AaB, [Bc/cA][AO/Oa][aC/CB] Multiply these equations together. Inverse as in Menelaus. 525. Corollary. If transversals through O from the vertices. of any odd polygon meet the sides AB, BC, CD, etc., in the points a, b, c, etc., in order, then [Aa/aB][Bb/bC][Cc/cD][Dd/dE] . . . etc. = 1. 526. Theorem. If any transversal cuts the sides of a triangle and their three intersectors AO, BO, CO, in the points. A', B', C', a', b', c', respectively, then [A'b'/b' C'][C'a'/a'B' [B'c'/c'A'] = 1. Proof. Each side forms a triangle with its intersector and Take the four remaining straights in succes the transversal. sion for transversals to each triangle, applying Menelaus symmetrically, and combine the twelve equations. 526 (b). The whole straight determined by two points may be called the join of the two points. The point determined by two straights, their intersection, may be called the cross of the two straights. 527. [Desargues.] If the vertices of two triangles join concurrently, the pairs of corresponding sides intersect co-straightly, and inversely. P R B Proof. Take bc, ca, ab, transversals respectively to the triangles OBC, OCA, OAB; apply Menelaus, and the product of the three equations shows that P, Q, R lie on a transversal to ABC. FIG. 234. 528. Corollary. For two figures such that to every point of one corresponds a point of the other, to every straight of the one a straight of the other, to every join of the one the join of the corresponding points of the other, to every cross of the one the cross of the corresponding straights of the other; when all the corresponding crosses join concurrently, then all the corresponding joins cross costraightly; and inversely. 529. The straight on which the pairs of sides cross is called the axis of perspective. The figures are called complete perspec tives. 530. The projection of a point on a sect is the foot of the perpendicular from the point to the straight of the sect. projection of a sect on a straight is the piece between the perpendiculars dropped upon the straight from the 531. The ends of the sect. 532. Theorem. The projections on the sides of a triangle of any point on its circumcircle are co-straight. [This straight is called the Simson's straight of the triangle with respect to the given point.] Proof. Let O be any point on circumcircle projections GF, GH. of ▲ ABC. Join its Join OB, OC. Since OGC and x OHC are B r't, therefore C, H, G, O are concyclic. Similarly, G, B, F, O are concyclic. :: ☀ OGF = ¥ OBF, inscribed angles on same arc of circle OGBF. FIG. 235. OBA. OCH = s't x. But & OBF = x OCA, being supplemental to .. OGH + OGF = 4 OGH + 533. Inverse. If the projections on the sides of a triangle of a point be co-straight, that point is on the triangle's circumcircle. Proof. Let G, H, F, 1 projections of O straight. Since O, C, H, G are concyclic, .. being supplements of But OGF= circle OGBF. OGH. on a, b, c be coOCH = 4 OGF, X CHAPTER II. HARMONIC RANGES AND PENCILS. 534. A system of co-straight points is called a range, of which the straight is the bearer. 535. A system of concurrent straights is called a pencil, of which the intersection point is the vertex or the bearer. 536. Straights all parallel form a pencil of parallels or a parallel-pencil. 537. Thus straights with equal perpendiculars from two given points form two pencils, one parallel to their join, and the other bisecting it. 538. A range and a pencil are called perspective when each point of the range lies on a straight of the pencil. 539. Two ranges are called perspective when their points lie in pairs on the straights of a pencil. The bearer of the pencil is called the perspective-center. 540. Two pencils are called perspective when their straights cross in pairs in the points of a range. The bearer of the range is called the projection axis. 541. Ranges and pencils are called projective if they can be put in perspective position. 542. If A, B be two points, and C, D, two co-straight with them, be so taken that AC/BC= AD/DB, then the points. A, C, B, D form a harmonic range; C and D are harmonic conjugates with respect to A and B: AC, AB, AD are said to be in harmonic progression: and AB is said to be a harmonic mean between AC and AD. Thus we have proved (479) that if C and D are harmonic conjugates with respect to A and B, then A and B are harmonic conjugates with respect to C and D. |