Book Y. ABC would be lefs than the angle ACB; but it is not; there, fore the fide AC is not less than AB; and it has been fhewn that it is not equal to AB; therefore AC is greater than AB. Wherefore the greater angle, &c. Q. E. D. b 18, I. See N. 3.5. 5. I. C 19. I. See N. PROP. XX. THEOR. NY two fides of a triangle are together greater than A the third fide. Let ABC be a triangle; any two fides of it together are greater than the third fide, viz. the fides BA, AC greater than the fide BC; and AB, BC greater than AC; and BC, CA greater than AB. Produce BA to the point D, and make AD equal to AC; and join DC. becaufe DA is equal to AC, the angle ADC is likewise equal b to ACD; but the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and be C B D cause the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater fide is oppofite to the greater angle; therefore the fide DB is greater than the fide BC; but DB is equal to BA and AC; therefore the fides BA, AC are greater than BC. In the fame manner it may be demonstrated, that the fides AB, BC are greater than CA, and BC, CA greater than AB. Therefore any two fides, &c. Q. E. D. PROP. XXI. THEOR. from the ends of the fide of a triangle, there be drawn two straight lines to a point within the triangle, thefe fhall be lefs than the other two fides of the tri angle, but fhall contain a greater angle. Let the two ftraight lines BD, CD be drawn from B, C, the ends of the fide BC of the triangle ABC, to the point D within it; BD and DC are lefs than the other two fides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC. Produce BD to E; and because two fides of a triangle are greater than the third fide, the two fides BA, AE of the tri angle angle ABE are greater than BE. To each of thefe add EC Book 1. therefore the fides BA, AC are greater than BE, EC: Again, because the two fides CE, ED of the triangle CED are greater than CD, add DB to each of thefe; therefore the fides CE, EB are greater than CD, DB; but it has been fhewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. B E Again, because the exterior angle of a triangle is greater than the anterior 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. PRO B. T Let A, B, C be the three given straight 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 thall be equal to A, B, C, each to each. Take a ftraight line DE terminated at the point D, but un limited towards E, and make DF equal to A, FG to B, and GH equal to C; and from the centre F, at the distance FD, defcribe the circle DKL; and D from the centre G, at the diftance GH, delibe another circle HLK, and join KF, KG; the triangle KFG has its fides equal to the three straight lines, A, B, C. Sce N! 2.20.1. a 3. I. G b. 3. Post, HE A B C Because the point F is the centre of the circle DKL, FD is equal Book I. equal to FK; but FD is equal to the ftraight line A: there fore FK is equal to A: Again, because G is the centre of the c 15. Def. circle LKH, 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 straight lines A, B, C. Which was to be done. a 22. I. b 8. 1. See N. A1 PROP. XXIII. PROB. Ta given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angie; it is required to make an angle at the given point A in the given straight line AB, that fhall be equal to the given rec tilineal angle DCE. Take in CD, CE, any points D, E, and join DE; and make a the triangle AFG the fides D. of which shall be equal to the three ftraight lines CD, DE, CE, fo AA B that CD be equal to AF, CE to AG, and DE to FG ; and becaufe DC, CE are equal to FA, AG, each to each, and the bafe DE to the bafe 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 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 base of that which has the greater angle fhall be greater than the base of the other. Let ABC, DEF be two triangles which have the two fides AB, AB, AC equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the bafe BC is alfo greater than the bafe EF. Book I. 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 ftraight line DE, make the angle EDG equal to the angle BAC; and make a 23, 1. DG equal to AC or DF, and join EG, GF. a 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 BAC is equal to the angle EDG;A D E b 3. I. C 4. I. d 5.1. G F e 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, and that the greater fide is oppofite to the greater angle; the fide EG e 19. 1. is therefore greater than the fide EF; but EG is equal to BC; and therefore alfo BC is greater than EF. Therefore, if two triangles, &c. Q. E. D. IF two triangles have two fides of the one equal to two fides of the other, each to each, but the bafe 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 base CB is greater than the bafe EF; the angle BAC is likewife greater than the angle EDF. For, Book I. # 4. I. b24.1. a A D 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 equal a to EF; but it is not; therefore the angle BAC is not equal to the angle EDF; neither is it lefs; becaufe then the base BC would be lefs b than the bafe EF; but it is not; therefore the B C E F angle BAC is not less than the angle EDF; and it was fhewn that it is not equal to it; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E.D. IF 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. either 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 also 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; also one fide equal to one fide; and first let those fides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF; the other A fides fhall be equal, For, if AB be not D equal to DE, one of B them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is equal |