BOOK SIXTH. Definitions. 1. Similar rectilineal figures are those which have their angles equal, each to each; and the sides about the equal angles proportionals. in 2. Reciprocal figures, viz. triangles and parallelograms, are those which have their sides about two of their angles proporti such manner, that the extremes belong to one of the figures, and the means to the other; and such figures are equal to one another. 3 A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment to the less. 4. The altitude of any figure is the distance between two parallel lines touching the figure, one of which is its base. Propositions. 1 Th. Triangles and parallelograms of the same, or equal altitudes, are to each other as their bases Given the triangles ABC, ACD, and the parallelograms CE, CF, on the bases BC, CD, and between two parallels (a); then ABC is to ACD, and CE is to CF, as BC is to CD. Take HC, CL, equimultiples of BC, CD; and divide CH, CL H into parts equal to CB, CD: then, because each of the parts GB, GH equals BC; and each of the parts KD, KL equals CD; therefore CH and CL are 3d multiples of BC and CD (6). To the point A join the points H, G, K, L: then the triangles ABC, AGB, AHG are all equal; and likwise the triangles ACD, ADK, AKL (c); and the triangles ACH, ACL are 3d multiples of the triangles ABC, ACD: and so, if the base CH be greater than the base CL, the triangle ACH is greater than the triangle ACL; if equal, equal; and if less, less. (d). Now the bases BC, CD, and the triangles ABC, ACD, are four magnitudes; and of the 1st BC, and 3d ABC, equimultiples CH, ACH, are taken; and of the 2d and 4th CD, ACD, equimultiples CL, ACL, are taken; also it is shown, that if the base CH be greater than the base CL, the triangle ACH is greater than the triangle ACL; if equal, equal; and if less, less: therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD (d). But the parallelograms CE, BF are 2d multiples of the triangles ABC, ACD (e), and the equimultiples have the ratio of their magnitudes (f); therefore, ABC:ACD::CE: CF. And, since ABC: ACD::BC:CD; therefore (g) also CE: CF::BC:CD. Wherefore, triangles, &c. Recite (a) def. 4, 6; Q. E. D. (d) def. 5, 5; (g) p. 11, 5. Cor. It may be also proved (in the same way,) that triangles and parallelograms of the same or equal bases are to each other as their altitudes. 2 Th. If a straight line be drawn parallel to one side of a triangle, it shall cut the other sides, or those sides produced proportionally: and if the sides, or the sides produced, be cut proportionally, the straight line joining the sectional points, shall be parallel to the third side of the triangle. Given the triangle ABC, and DE drawn parallel to BC, cutting AB, AC; 1, within the triangle; 2, below the base; 3, above the vertex: in either case, BD: AD:: CE: AE. Join BE, CD. 1. The triangles BDE, CDE are equal; D being on the same base DE, and between the same parallels BC, DE (a): but ADE is another triangle, to which the equals BDE, CDE havethe same ratio (6); and BDE: ADE:: BD: AD (c); also, CDE: ADE:: CE: AE (c): therefore BD is to E B C B C D E D AD as CE is to AE (d). 2. Upon the same construction. If BD: AD :: CE: AE, then DE is parallel to BC. For BD: AD :: BDE: ADE; and CE: AE:: CDE: ADE (c): therefore, BDE and CDE having the same ratio to ADE are equa' (b); and being on the same base DE, and same side of it, they are also between the same parallels (e): therefore DE is parallel to BC. Wherefore, if a straight line, &c. Recite (a) p. 37, 1; (d) p. 11, 5; R Q. E. D. (b) p. 7, 9, of b. 5; (c) p. 1, 6; C 3 Th. If a straight line bisect an angle of a triangle, and also cut the base; the segments of the base shall have the same ratio as the sides which contain the angle: and if the segments of the base have the same ratio as the other sides, the straight line drawn from the vertex shall bisect the vertical angle. Given the triangle ABC, and AD bisecting the angle A: then BD:DC::BA: AC;--and, if so, the angle A is bisected by AD. Through C draw CE parallel to AD (a); and produce BA to meet CE in E. A E A 1. Because BE and AC meet the parallels AD, CE, the exterior angle BAD, and the interior BEC, or AEC are equal (6), also the alternate angles DAC, ACE (c): but BAD equals DAC, by hypothesis; therefore AEC equals ACE (d), and the side AE equals the side AC (e). And because AD is drawn parallel to CE; therefore, BD:DC:: BA: AE, (f), or its equal AC. B D C 2. Because BD:DC::BA: AC, by hyp. and, that from the parallels, BD: DC::BA:AE(f); therefore AE equals AC (g); and their opposite angles AEC, ACE are equal (h): but AEC equals BAD (6), and DAC equals ACE; therefore BAD equals DAC (1), and so, the angle BAC is bisected by AD. Wherefore, if a straight line bisect, &c. Q. E. D. Recite (a) p. 31, 1; (b) p. 29, 1; (d) ax. 1, 1; (e) p. 6, 1; (c) p. 27, 1; (g) p. 9, 5; (h) p. 5, 1; (i) ax. 1, 1. A Th. If a straight line bisect the exterior angle of a triangle and meet the base produced, the segments between the meeting point and each extremity of the base shall have the same ratio as the other sides of the triangle. And if the segments of the base produced have the same ratio as the other sides, the straight line drawn from the vertex bisects the exterior angle. Given the triangle ABC, its side BA produced to E, the exterior angle CAE bisected by AD which meets BC produced in D. BD: DC::BA:AC; and if so, the exterior angle CAE is bisected by AD. E A F C D Through C draw CF parallel to AD (a). 1. Because the straight lines AC, FE meet the parallels AD, CF, the alternate angles DAC, В ACF are equal (6), and so are the interior and exterior AFC, DAE (c): but by hyp. DAC equals DAE; therefore ACF equals AFC (d), and AF equals AC (e): and because CF is drawn parallel to AD; therefore BD: CD:: BA: AF (f), or AC its equal (f). 2. Again, because BD: DC:: BA: AC, or AF; therefore AC equals AF (g), and the angles AFC, ACF are equal (h); but the interior angle AFC equals the exterior DAE, and the alternate angles ACF, DAC are equal (b); therefore DAE equals DAC (d); and so, the angle CAE is bisected by AD. Wherefore, if one side of a triangle be produced, &c. Q. E. D. 4 Th. The sides about the angles of equiangular triangles are proportionals; and the sides which are opposite to equal angles are homologous; that is, they are all antecedents, or all consequents of the ratios. Given two triangles ABC, DCE, equiangular at Band C, at Cand E, and therefore at A and D (a); the sides adjacent to these equal angles are proportionals, and the sides opposite to them are homologous. A D C E Place the bases BC, CE in the same straight line; and because the angles ABC, ACB are less than two right angles (b), their equals ABC, DEC B are also less, and BA, ED produced will meet in some common point F; and because the angles ABC, DCE are equal, BF is parallel to CD (c); also, because ACB equals DEC, AC is parallel to FE; therefore FACD is a parallelogram, whose opposite sides AC, DF, and AF, CD are equal (d). And, in the triangle EBF, because AC is parallel to FE, and CD parallel to BF; therefore, BA: AF, or its equal CD :: BC: CE, and BC: CE:: FD, or its equal AC: DE (e). And taking both these alternately (f), BC:CA::CE: ED; and again ex æquo; BA:AC::CD: DE, (g). Wherefore, the sides about the angles, &c. Q. E. D. 5 Th. If the sides of two triangles about each of their angles be proportionals, the triangles shall be equiangular, and shall have their equal angles opposite to homologous sides. that Given two triangles ABC, DEF; so A D æquali AB: CA::DE:FD; B At the points E, F, in the straight line EF, make the angles FEG, EFG severally equal to the angles B, C(a); then the third angles G and A are equal (6) Now, because the triangles ABC, EGF are equiangular, their sides about equal angles are proportionals (c); wherefore AB: BC::GE: EF; but AB: BC:: DE: EF, by hyp., and so, DE: EF::GE: EF, and therefore GE equals DE (d). For the same reason GF equals DF. And since, in the triangles DEF, GEF, the sides EG, ED are equal, and EF is common, there are two sides in the one equal to two sides in the other; and the bases FG, FD are equal; therefore the angles DEF, GEF are equal; as are also the other angles, namely, DFE to GFE and D to G (e). And because DEF equals GEF, it also equals B (f): for the same reason the angle C equals DFE, and the angle A equals the angle D. Therefore the triangles ABC and DEF are equiangular. If therefore, the sides of two triangles, &c. Q. E. D. Recite (a) p. 23, 1; (d) p. 9, 5; (b) 32, 1; (c) p. 4, 6; (f) ax. 1, 1. 6 Th. If two triangles have one angle of the one equal to one of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite homologous sides. Given two triangles ABC, DEF, having the angles A and D equal, and BA to AC as ED to DF; then shall the other angles be equal; namely, B to E and C to F. A D G At the points D, F, in the side DF, make the angle FDG equal to FDE, or A, and DFG equal to C (a); the third angles G and B are equal (6). EF B Now, since the triangles ABC, DGF are made equiangular, BA:AC::GD: DF (c); but BA: AC:: ED: DF by hyp.; therefore ED: DF:: GD: DF (d); and so, ED equals GD, (e), and DF is common to the two triangles EDF, GDF; therefore the sides ED, DF equal the sides GD, DF, and they contain equal angles: where |