BCA; wherefore the remaining angle BAC is equal to the remaining angle EGF (32. 1.), and the triangle ABC is therefore equiangular to the triangle GEF; and consequently they have their sides opposite to the в equal angles proportionals (4. 6.). Wherefore, as AB to BC, so is GE to EF; but as AB to BC, SO is DE to EF; therefore as DE to EF, so (11. 5.) GE to EF: therefore DE and GE have the same ratio to EF, and consequently are equal (9. 5.): for the same reason, DF is equal to FG: and because in the triangles DEF, GEF, DE is equal to EG, and EF common, the two sides DE, EF are equal to the two GE, EF, and the base DF is equal to the base GF: therefore the angle DEF is equal (8. 1.) to the angle GEF, and the other angles to the other angles which are subtended by the equal sides (4. 1.): wherefore the angle DFE is equal to the angle GFE, and EDF to EGF: and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF: for the same reason the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D. PROP. VI. THEOR. If two triangles have one angle of the one equal to one angle 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 to the homologous sides. Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE. At the points D, F, in the straight line DF, make (23. 1.) the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB; wherefore A the remaining angle at B is equal to the remaining one at G (32. 1.), and, consequently the triangle ABC is equiangular to the triangle DGF; and therefore as BA to AC, so is (4. 6.) GD to DF: but, by the hypothe! sis, as BA to AC, so is ED to DF; I as therefore ED to DF, so is (11.5.) B C E F GD to DF; wherefore ED is equal (9. 5.) to DG; and DF is common to the two triangles EDF, GDF; therefore the two sides ED, DF are equal to the two sides GD, DF: and the angle EDF is equal to the angle GDF; wherefore the base EF is equal to the base FG (4. 1.), and the triangle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides; therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E: but the angle DFG is equal to the angle ACB; therefore the angle ACB is equal to the angle DFE: and the angle BAC is equal to the angle EDF (Hyp.); wherefore also the remaining angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D. PROP. VII. THEOR. 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 have those angles equal about which the sides are proportionals.* 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 is equiangular to the triangle DEF, viz. the angle ABC is equal to the angle DEF, and the remaining angle at C, to the remaining angle at F. For, if the angles ABC, DEF be not equal, one of them is greater than the other: let ABC be the greater, and at the point B, in the straight line AB, make the angle Α ABG equal to the angle (23. 1.) DEF: and because the angle at A is equal to the angle at D, and the angle ABG to the angle DEF; the remaining angle AGB is equal B CEF (32. 1.) to the remaining angle DFE: therefore the triangle ABG is equiangular to the triangle DEF; wherefore, (4. 6.) as AB is to BG, so is DE to EF; but as DE to EF, so, by hypothesis, is AB to BC; therefore as AB to BC, so is AB to BG (11. 5.); and because AB has the same ratio to each of the lines BC, BG; BC is equal (9. 5.) to BG, and therefore the angle BGC is equal to the angle BCG (5. 1.); but the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle, and D * See Note. the adjacent angle AGB must be greater than a right angle (13. 1.). 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 hypothesis, 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 equal to the angle at D; wherefore the remaining angle at C is equal to the remaining angle at F: therefore the 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 is also in this case equiangular to the triangle DEF. The same construction being made, it may be proved in like D manner that BC is equal to BG, and the angle at C equal to the G . angle BGC: but the angle at C is not less than a right angle ; therefore the angle BGC is not less B C E F than a right angle; wherefore two angles of the triangle BGC are together not less than two right angles, which is impossible (17. 1.): and therefore the 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 is equiangular to the triangle DEF. For, if they be not equiangular, make, at the point B of the straight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC; but the angle BCG is a right angle, therefore (5. 1.) the angle BGC is also a right angle; whence two of the angles of the triangle BGC are together not less than two right angles, which is impossible (17. 1.); therefore Basic the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D. PROP. VIII. THEOR. 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 triangle, and to one another. * * See Note. Let ABC be a right angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC: the triangles ABD, ADC are similar to the whole triangle ABC, and to one another. Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD (32. 1.): therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals (4. 6.); wherefore the triangles are [ Ꭰ C similar (1. Def. 6.); in the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC: and the triangles ABD, ADC, 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 its segment adjacent to that side: because in the triangles BDA, ADC, BD is to DA as DA to DC (4. 6); and in the triangles ABC, DBA, BC is to BA, as BA to BD (4. 6.); and in the triangles ABC, ACD, BC is to CA as CA to CD (4. 6.). PROP. IX. PROB. From a given straight line to cut off any part required.* Let AB be the given straight line; it is required to cut off any part from it. 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 A from it: join BC, and draw DE parallel to it: then AE is the part required to be cut off. Because ED is parallel to one of the sides of the E Led triangle ABC, viz, to BC, as CD is to DA, so is (2. 6.) BE to EA ; and, by composition (18. 5.) CA is to AD as BA to AE: but CA is a multiple of AD; therefore (D. 5.) BA is the same multiple of AE: whatever part therefore AD is of AC, AE is 1 the same part of AB: wherefore, from the straight. line AB the part required is cut off. Which was to be done. *Sec Note. PROP. X. PROB. To divide a given straight line similarly to a given divided 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 (31. 1.) DF EG parallels to it; and through D draw DHK parallel to AB: therefore each of the figures FH, HB is a parallelogram; wherefore DH is equal (34. 1.) to FG, and HK to GB: and because HE is parallel to KC, A one of the sides of the triangle DKC, as CE to ED, so is (2. 6.) KH to HD: but KH is equal to BG, and HD to GF; therefore as CE to ED, so is BG to GF; again, because F FD is parallel to EG, one of the sides of the triangle AGE, as ED to DA, so is GF to a FA; but it has been proved that CE is to ED as BG to GF; and as ED to DA, so GF B K to FA: therefore the given straight line AB is divided similarly to AC. Which was to be done. G-HE PROP. XI. PROB. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; it is required to find a third proportional to AB, AC. Produce AB, AC to the points D, E: and make A BD equal to AC; and having joined BC, through D draw DE parallel to it (31. 1.). Because BC is parallel to DE, a side of the triangle ADE, AB is (2. 6.) to BD, as AC to CE: but B BD is equal to AC; as therefore AB 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. PROP. XII. PROB. 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. |