4 32. 1. & ing angle at B is equal t to the remaining angle at E: therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. 3 Ax. Q. E. D. PROP. VII. THEOR. See N. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals; then, if each of the remaining angles be either less, or not less, than a right angle, or if one of them be a right angle ; the triangles shall be equiangular, and shall have those angles equal about which the sides are proportionals. A D Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz, the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF; and, in the first case, let each of the remaining angles at C, F be less than a right angle. The triangle ABC shall be equiangular to the triangle DEF, viz. the angle ABC shall be equal to the angle DEF, and the remaining angle at C equal to the remaining angle at F. For, if the angles ABC, DEF be not equal, one of them must be greater than the other : let ABC be the * 23.,1. greater, and at the point B, in the straight line AB*, make the angle ABG equal to the angle DEF; and because the angle at A is 4 Hyp. equalt to the angle at D, and the angle ABG to the angle DEF; the remain* 32. 1.& ing angle AGB is equal* to the remain ing angle DFE: therefore the triangle ABG is equiangular to the triangle DEF: wherefore * as AB is to BG, so is DE to EF: but as DE to EF so, by hypothesis, is * 11.5. AB to BC: therefore* as AB to BC, so is AB to BG: and because AB has the same ratio to each of the lines BC, BG; BC is equal* to BG; and therefore the angle BGC is equal * to the angle BCG: but the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle; and therefore the adjacent angle AGB must be greater than a right angle: but it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle: but, by the hypo G B 4. 6. * 9.5. * 5. 1. * 13, 1. 3 Ax. A D B E thesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal, that is, they are equal : and the angle at A is equalt to the + Hyp. angle at D; wherefore the remaining angle at C is equal † to the 'remaining angle at F: therefore the + 32. 1. & triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle: the triangle ABC shall also in this case be equiangular to the triangle DEF. The same construction being made, it may be proved in like manner that BC is equal to BG, and therefore the angle at C equal to the angle BGC: but the angle at C is not less than a right angle: therefore the angle + Hyp. BGC is not less than a right angle: wherefore two angles of the triangle BGC are together not less than two right angles ; which is impossible*; and therefore the * 17. 1. triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle: in this case likewise, the triangle ABC shall be equiangular to the triangle DEF. For, if they be not equiangular, at the point B in the straight line AB make the angle ABG equal to the angle DEF: then it may be D proved, as in the first case, that BG is equal to BC: and therefore * the E angle BCG equal to the angle BGC: but the angle BCG is a right + angle, therefore + the angle + Hyp. BGC is also a right angle; whence two of the angles + 1 Ax. of the triangle BGC are together not less than two right angles; which is impossible *: therefore the tri- . 17. 1. angle ABC is equiangular to the triangle DEF. Wherefore if two triangles, &c. B * 5. 1. B G Q. E. D. PROP. VIII. THEOR. from the right angle to the base; the triangles on each B D C 3 Ax. * 4. 6. 1 Def. 6. are right angle BAC; and from the point A let AD be drawn perpendicular to the base BC: the triangles ABD, ADC shall be similar to the whole triangle ABC, and to one another. Because the angle BAC is equal to the angle † 11 Ax. ADB+, each of them being a right the remaining angle ACB is equal to * 32. 1. & the remaining angle BAD*: there fore the triangle ABC is equiangular * 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 BD; and in the triangles ABC, ACD, BC is to CA *, as CA to CD. * 4. 6. * 4. 6. the 4. 6. PROP. IX. PROB. See N. any part from it. From a given straight line to cut off any part required. Let AB be the given straight line; it is required to cut off From the point A draw a straight line AC, making any angle with AB; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it; join BC, ELD and draw DE parallel to it: then AE shall be the part required to be cut off. Because ED is parallel to one of the B sides of the triangle ABC, viz, to BC, as CD is to DA so is* BE to EA; and by composition*, . 2.6. • 18.5. CA is to AD, as BA to AE: but CA is a multiplet + Constr. of AD; therefore* BA is the same multiple of AE: D. 5. whatever part therefore 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. To divide a given straight line similarly to a given di vided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have. 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, E, draw* DF, EG .31. 1. 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 parallelogram: wherefore DH D is equal* to FG, and HK to GB: and HIE because HE is parallel to KC, one of the sides of the triangle DKC, as CE to ED, so is* KH to HD: but KH is equal to BG, • 2. 6. and HD to GF; therefore, as CE to EDt, so is BG +7.5. to GF: again, because FD is parallel to GE, one of the sides of the triangle AGE, as ED to DA, so ist + 2. 6. 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. 34. 1. B K PROP. XI. PROB. To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines : it is required to find a third proportional to AB, AC. Let AB, AC be placed so as to contain any angle: L 1 A * 31. 1. B В produce AB, AC, to the points_D, E; D E * 2. 6. 17. 5. PROP. XII. PROB. + 3. 1. D * 31.1. B с H G To find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines; it is required to find a fourth proportional to A, B, C. Take two straight lines DE, DF, containing any A Because GH is parallel to EF, of the sides of the triangle DEF, DG is to GE*, as DH to HF; but DG is equal to A, GE to B, and DH to C; therefore, as A is to Bt, so is C to HF. Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done. PROP. XIII. PROB. E To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines : it is required to find a mean proportional between them. 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 the в с angle ADC in a semicircle is a right 11. 1. |