are greater than BE. to each of these add EC, therefore the fides Book I. than BE, EC; much more then are BA, AC greater than BD, DC. Again because the exterior angle of a triangle is greater than the interior and oppofite angle, the exterior angle BDC of the triangle CDE is greater than CED. for the fame reason, the exterior angle CEB of the triangle ABE is greater than BAC. and it has been demonftrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. therefore if from the ends of, &c. Q. E. D. PROP. XXII. PROB. O make a triangle of which the fides fhall be equal to three given straight lines; but any two whatever of these must be greater than the thirda. Let A, B, C be the three given ftraight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the fides shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unlimited See N. a. 20. I. has its fides equal to the three ftraight lines A, B, C. Because the point F is the center of the circle DKL, FD is e e Book I. qual to FK; but FD is equal to the straight line A; therefore FK is equal to A. again, because G is the center of the circle LKH, c. 15. Def. GH is equal to GK; but GH is equal to C, therefore alfo GK is equal to C. and FG is equal to B; therefore the three straight lines KF, FG, GK are equal to the three A, B, C. and therefore the triangle KFG has its three fides KF, FG, GK equal to the three given ftraight lines A, B, C. Which was to be done. a. 22. I. b. 8. I. See N. PROP. XXIII. PROB. Ta given point in a given ftraight line to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given ftraight line, and A the given point in it. and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given ftraight line AB that fhall be equal to the given rectilineal angle DCE. AA B to the three fuaight lines CD, DE, EC, so that CD be equal to AF, CE to AG, and DE to FG. and because DC, CE are equal to FA, AG, each to each, and the bafe DE to the base FG; the angle DCE is equal to the angle FAG. therefore at the given point A in the given ftraight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROP. XXIV. THEOR. IF F two triangles have two fides of the one equal to two fides of the other, each to each, but the angle contained by the two fides of one of them greater than the angle contained by the two fides equal to them, of the other; the bafe of that which has the greater angle shall be greater than the bafe of the other. Let ABC, DEF be two triangles which have the two fides AB, AC AC equal to the two DE, DF, each to each, viz. AB equal to DE, Book I. and AC to DF; but the angle BAC greater than the angle EDF. the base BC is alfo greater than the base EF. Of the two fides DE, DF let DE be the fide which is not greater than the other, and at the point D in the straight line DE make a. 23. 1. the angle EDG equal to the angle BAC; and make DG equal bb. 3. 1. to AC or DF, and join EG, GF. Because AB is equal to DE, and AC to DG, the two fides BA, AC are equal to the two ED, DG, each to each, and the angle C. 4. I. d. 5. 1. fore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF. and because the angle EFG of the triangle EFG is greater than its angle EGF, e and that the greater fide is oppofite to the greater angle; the e. 19. 1, fide EG is therefore greater than the fide EF. but EG is equal to BC; and therefore alfo BC is greater than EF. therefore if twą triangles, &c. Q. E. D. I PROP. XXV. THEOR. F two triangles have two fides of the one equal to two fides of the other, each to each, but the base of the one greater than the bafe of the other; the angle alfo contained by the fides of that which has the greater base, shall be greater than the angle contained by the fides equal to them, of the other. Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the bafe CB greater than the base EF. the angle BAC is likewise greater than the angle EDF. Book I. 2.4. 1. For if it be not greater, it must either be equal to it, or lefs. but the angle BAC is not equal to the angle EDF, because then the base BC would be b. 24. 1. b than the bafe EF; B C E F the angle BAC is not lefs than the angle EDF. and it was fhewn that it is not equal to it; therefore the angle BAC is greater than the angle LDF. Wherefore if two triangles, &c. Q. E. D. IF PROP. XXVI. THEOR. F two triangles have two angles of one equal to two angles of the other; each to each, and one fide equal to one fide, viz. cither the fides adjacent to the equal angles, or the fides oppofite to equal angles in each; then fhall the other fides be equal, each to each, and alfo the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD; alfo one fide equal to one fide; and first, let thofe fides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF. the other fides shall be equal, each to each, viz. AB to DE, and AC to DF; and the third angle BAC to G For if AB be not equal to DE, one of A D and make BG equal to DE, and join GC. therefore because BG is equal a qual to DE, and BC to EF, the two fides GB, BC are equal to the Book I. two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the bafe GC is equal to the bafe DF, and the tri-a. 4. 1. angle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal fides are oppofite; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothefis, equal to the angle BCA; wherefore alfo the angle BCG is equal to the angle BCA, the less to the greater, which is impoffible. therefore AB is not unequal to DE, that is, it is equal to it. and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF, the bafe therefore AC is equal to the base DF, and the third angle BAC to the third angle EDF. Next, let the fides which are oppofite to equal angles in each triangle be equal to one another, viz. AB to DE; likewife in this cafe, the other fides fhall be equal, 2 D For if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH. and becaufe BH is equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles fhall be equal, each to each, to which the equal fides are oppofite. therefore the angle BHA is equal to the angle EFD. but EFD is equal to the angle BCA; therefore also the angle. BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and oppofite angle BCA; which is impoffibleb. wherefore BC is not unequal to EF, that is, b. 16. 1. it is equal to it; and AB is equal to DE; therefore the two AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the bafe AC is equal to the base DF, and the third angle BAC to the third angle EDF. therefore if two triangles, &c. Q.E. D. PROP. |