mon and * 4.6. * Because the angle BAC is equal to the angle +11 Ax. ADB,+ each of them being a right angle, and that the angle at B is com A angle ACB is equal to the *32. 1. remaining 'angle BAD : * B D C therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals;* *! Def.6. wherefore the triangles are * similar: in the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC. † And the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right angled, &c. Q. E. D. Cor. From this it is manifest that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base, and the segment of it adjacent to that side : because in the triangles BDA, ADC, BD is to DA, * as DA to DC, and in the triangles ABC, DBA, BC is to BA, as BA to and in the triangles ABC, ACD, BC is to СА, as CA to CD. *4 6. *4.6. * BD; *4.6 PROP. IX. PROB. required. From the point å draw a straight line AC, † The triangles ADB, ADC, being both equiangular to † 1 Ax. ABC, are equiangular + to each other; therefore they have † 4.6. the sides about their equal angles † proportionals; therefors t1Def.6, they are similar to each other. making any angle with AB; and in AC take E Because ED is parallel to one of the sides of the triangle ABC, viz. B to BC, as CD is to DĂ, so is * BE to EA; and * 2.6. hy composition, * CA is to AD, as BA to AE : *18. 5. but CA is a multiple t of AD; therefore * BA is +Const. the same multiple of AE: whatever part there- * D.5. fore AD is of AC, AE is the same part of AB: wherefore, from the straight line AB the part required is cut off. Which was to be done. PROP. X. PROB. given divided straight line, that is, into parts Let AB be the straight line given to be divided, and AC the divided line : it is required to divide AB similarly to AC. Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, É draw* *31. 1. DF, EG parallels to it: AB shall be divided in the points F, G similarly to AC. Through D draw DHK parallel to AB: therefore each of the figures, FH, HB, is a parallelo * FU gram; wherefore DH is equal * GI HAE to FG, and HK to GB: and because HE is parallel to KC, K one of the sides of the triangle DKC, as CE to ED, so is * KH to HD: but * 2.6. KH is equal to BG, and HD to GF; therefore, * 34. 1. B R +7.5. + 2.6 as CE to ED,+ so is BG to GF: again, because FD is parallel to GE, one of the sides of the triangle AGE, as ED to DA, so ist GF to FA: but it has been proved that CE is to ED, as BG to GF: therefore as CE is to ED, so is BG to GF, and as ED to DA, so GF to FA : therefore the given straight line AB is divided similarly to AC. Which was to be done. PROP. XI. PROB. lines. Let AB, ACbe placed so as to contain any angle: produce AB, AC to the points D, E; and make BD equal to AC; A proportional to AB and AC. D E to BD, as AC to CE: but BD is equal to AC; 77. 5. therefore as AB is to + AC, so is AC to CE. Wherefore, to the two given straight lines AB, AC, a third proportional CE is found. Which was to be done. a * 2.6. PROP. XII. PROB. straight lines. Take two straight lines DE, DF, containing +& 1 any angle EDF; and upon these † make DG equal to A, GE equal to B, and DH equal to C; join GH, and through E draw B. EF parallel * to it: HF ¢ *31. 2. shall be a fourth propor- G II tional to A, B, C. Because GH is parallel E to EF, one of the sides of the triangle DEF, DG is to GE,* as DH to *2. & HF; but DG is equal to A, GE to B, and DH to C; therefore, as A is to Bit so is C to HF. + 7. 5. Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done. * 1L PROP. XIII. PROB. straight lines. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw * BD at right angles to AC: BD shall be a mean proportional between AB and BC. Join AD, DC: and because А B C the angle ADC in a semicircle is a right angle,* and because in the right angled * 31. 3. triangle ADC, BD is drawn from the right angle perpendicular to the base, DB a mean proportional between AB, BC the segments of the base: * therefore between the two given straight .Cor.& lines AB, BC, a mean proportional DB is found. f. Which was to be done. PROP. XIV. THEOR. Equal parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional : and parallelograms that have one angle of the one equal to one angle of the other, and their about the equal angles reciprocally proportional, are equal to one another. Let AB, BC be equal parallelograms, which have the angles at B equal: the sides of the parallelograms AB, BC about the equal angles, shall be reciprocally proportional ; that is, DB shall be to BE, as GB to BF. Let the sides DB, BE be placed in the same straight line; wherefore also I FB, BG are in one * 14. 1. straight line: * complete the parallelogram FE: and because the parallelogram E * 7.5. AB is to FE, as BC to FE:* D G с to FE, so is the base GB to BF; therefore, as *11.5. DB to BE, so is GB to BF.* Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional. Next, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF: the parallelogram AB shall be equal to the parallelogram BC. Because, as DB to BE, so is GB to BF; and * * 1.6. + Hyp. 1 Because the angle DBF is equal † to the angle GBE, add to each the angle FBE: therefore the two angles DBF, +2 Ax. FBE are together equal + to the two angles FBE, EBG: but +13. 1. DBF, FBE are together equal † to two right angles; there. 11 Axl fore FBE, EBG are together equal † to two right angles : 14.1. therefore FB, BG are in the same* straight line. |