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. Recite (a) p. 31, 1; (d) ax. 1, 1; (g) p. 9, 5; 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 B and C, at C and 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 F 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 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 And taking both these alternately (ƒ), BA: BC: CD: CE; also BC: CA::CE: ED; and again ex æquo; 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. Given two triangles ABC, DEF; so that AB: BC: DE: EF, BC: CA: EF: FD, and ex æquali AB: CA:: DE: FD; then the triangles are equiangular, having their equal angles opposite to homologous sides 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 (b) 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 (ƒ): 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; (e) p. 8, 1; (c) p. 4, 6; 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. 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 (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 fore (f) the bases EF, GF are equal; the angles E, G are equal; also DFE equals DFG. But G is made equal to B, and DFG to C; therefore B is equal to E, and C to DFE (g). Wherefore, if two triangles have, &c. Recite (a) p. 23, 1; (d) p. 11, 5; Q. E. D. (b) p. 32, 1; (e) 9, 5; (c) p. 4, 6: 7 Th. If two triangles have two angles equal, and the sides about other two angles proportionals; then, whether the third two be oblique or right angles, the triangles shall be equiangular, and have those angles equal which are contained by the proportional sides. Let two triangles ABC, DEF, have equal angles at A and D, and have AB to BC as DE to EF; their angles at B and E are equal, also those at C and F. If then it be said, that the angles at B and E are unequal, make the angle ABG equal to E (d); and it follows, that AB: BG:: DE: EF (b); that AB has to BC and BG the same ratio; that BG equals BC (c), and that the angles BGC and BCG are equal (¿` Then since the angles at C and F may be right or oblique: 1. If C and F be acute, BGC is acute (d), and the adjacent angle BGA is obtuse (e); and so, F is obtuse, because the triangles ABG, DEF assume to be equiangular. 2. If C and F be obtuse, BGC is obtuse (d), and the adjacent angle BGA is acute (e); and So, F is acute, because the triangles ABG, DEF assume to be equiangular. 3. If C and F be right angles, BGC is a right angle (d) and so two angles of a triangle are not less than two right angles (ƒ). Therefore to deny the equal- B ity of the angles at B and E, A makes the same angle F acute and obtuse, and two angles of a trian gle equal to two right angles, which are both absurd. Wherefore, if two triangles have two angles equal, &c. Q. E. D. Recite (a) p. 23, 1; (d) p. 5, 1; (b) p. 4. 6; (e) p. 13, 1; (c) p. 9, 5; 8 Th. In a right angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole and to each other. Given the triangle ABC, right angled at C: the perpendicular CD divides ABC into two triangles similar to the whole and to each other. Because the angle BCA equals BDC (a), and that the angle B is common to the two triangles BCA, BDC, the third angles BAC, BCD, are equal to each other (b); therefore the triangles BCA and BDC are equiangular, and the sides about the equal A angles are proportionals (c): therefore the triangles are similar (d). B In like manner, it may be shown, that the triangles ADC, BDC are equiangular and similar: and the triangles BDC, ADC, being each equiangular and similar to BCA, are equiangular and similar to each other. Therefore, in a right angled triangle, &c. Q. E. D. Recite (a) ax. 10, 1; (d) def. 1, 6. (b) p. 32, 1; (c) p. 4, 6; Cor. From the equiangular triangles, as above, come the following proportions namely, BD: BC:: BČ: BA, AD: AC AC: AB, and BD: DC :: CD: DA. Therefore each side of the triangle ABC is a mean proportional between its adjacent segment of the base and the base complete; and the perpendicular is a mean proportional between the segments of the base 9 P. From a given straight line (AB), to cut off any part required. From the point A draw AC, at any angle with AB. In AC take a point D, and make AC the same multiple of AD that AB is of the part to be cut off from it: join BC, and draw DE parallel to it; AE is the part required. Because ED is parallel to BC, a side of the triangle ABC (a), CD: DA:: BE: EA; and by composition, CA: AD: : BA: AE (b): but CĂ is a multiple of AD, and BA is the same multiple B of AE (c): therefore, whatever part AD is of ÁC, the same part is AE of AB. D Wherefore, from the straight line AB the required part is cut off; which was to be done. Recite (a) p. 2, 6; (b) p. 18, 5; (c) p. D, 5, 10 P. To divide a given straight line into parts, having the same ratios as the parts of a divided straight line given. Given the straight line AC, divided in D, E; and AB to be divided into similar parts, or having the same ratios. Place AB, AC, so as to contain any angle; join BC; and through the points D, E draw DF, EG parallel to BC (a); and through D draw DHK parallel to AB. E D H Now, in the parallelograms FH, HB, the side DH is equal to FG, and HK to GB (b). And because, in the triangle DKC, HE is parallel to KC, CE is to ED as KH is to HD, or as BG is to GF and because, in the triangle AGE, FD is parallel to GE, ED is to DA as GF is to FA (c). And since it proves that CE: ED:: BG: GF, and ED: DA:: GF: FA; K Therefore, the given straight line AB is divided similarly to AC; which was to be done. Recite (a) p. 31, 1; (b) p. 34, 1; (c) p. 2, 6. 11 P. To find a third proportional to two given straight lines, AB, AC. Place the given lines AB, AC so as to contain any angle; join BC; produce AB, so that BD shall equal AC; through D, draw DE parallel to BC (a), and produce AC to meet DE (6). Because, in the triangle ADE, BC is parallel to DE, we have AB to BD, or to its equal AC, as AC to CE (c). Therefore, CE is the third proportional to AB, AC; which was sought. D B Recite (a) p. 31, 1; (b) pos. 2; (c) p. 2, 6. 12 P. To find a fourth proportional to three given straight lines A, B, C. Place two straight lines DE, DF so as to contain any angle EDF; and upon these make DG=A, GE=B, and DH=C; join GH, and produce DH to meet EF drawn parallel to GE. (a). And because, in the triangle DEF, GH is parallel to EF, there is DG to GE as DH to HF (b): but DG, GE, DH were made equal to A, B, C : C E Therefore, A: B::C: HF; and so, HF is the fourth proportional |