greater than the angle BCA; the side AC shall be greater than the side AB. For, if it be not greater, AC must either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal to the angle ACB; but it * 5. 1. is not; therefore AC is not equal to AB: neither is it less; because then the angle ABC greater than AB. PROP. XX. THEOR. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle: any two sides of it together shall be greater than the third side, viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and BC, CA greater than AB. In BA produced take AD equal to AC; and join DC. B C 18. 1. 5. 1. Because AD is equal to AC, the angle ADC is equal to ACD; but the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater angle is subtended by the greater side; * 19. 1. therefore the side DB is greater than the side BC; but BD is equal to BA and AC, because AC is equal to AD; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. Therefore any two sides, &c. Q. E. D. PROP. XXI. THEOR. If from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle; BD and DC shall be less than the two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC. A D Produce BD to meet AC in E; and because two sides of a triangle are greater than the third side, the two sides BA, AE of the triangle ABE are greater than BE. To each of these unequals add EC; therefore the sides BA, AC are greater than BE, EC. Again, because the two sides CE, ED of the triangle CED are greater than CD, add DB to each of these unequals; therefore the sides CE, EB are greater than CD, DB; but it has been shown that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. B Again, because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated 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. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third.* 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 sides shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unlimited towards E, and make* DF equal K *20. 1. 3. 1. at the distance GH, describe another circle HLK; 3 Post. and join KF, KG; the triangle KFG shall have its sides equal to the three straight lines, A, B, C. Because the point F is the centre of the circle DKL, FD is equal to FK; but FD is equal to the straight * 15 Def. line A; therefore FK is equal to A. Again, because G is the centre of the circle LKH, GH is equal to GK; * 15 Def. but GH is equal to C; therefore also GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are respectively equal to the three A, B, C: and therefore the triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines A, B, C. Which was to be done. PROP. XXIII. PROB. At a given point in a given straight line, to make a rec- D point in it, and DCE the given rectilineal angle; it is In CD, CE, take any points D * 22. 1. make* the triangle AFG the *8. 1. sides of which shall be equal to the three straight lines CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG; then shall the angle FAG be equal to the given angle C. Because FA, AG are equal to DC, CE, each to each, and the base FG to the base DE; therefore the angle FAG is equal to the angle DCE. Wherefore, at the given point A in the given straight 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 sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other. Let ABC, DEF be two triangles, which have the two sides 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 base BC shall be greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in 23. 1. the straight line DE, make the angle EDG equal to the angle BAC; and make DG equal to AC or DF, and join EG, GF. * 3. 1. * Because DE is equal to AB, and DG to AC, the two sides ED, DG are equal to the two BA, AC, each to each, and the angle EDG is equal to the angle BAC; therefore the base EG is equal to the base BC; and * because DG is equal to DF, the angle DFG is equal* to the angle DGF; but the B E 4. 1. * 5.1. angle DGF is greater than the angle EGF; therefore the angle DFG is greater than the angle EGF; much more then 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* angle * 19. 1. is subtended by the greater side; therefore the side EG is greater than the side EF; but EG has been proved to be equal to BC; and therefore BC is greater than EF. Therefore, if two triangles, &c. Q. E. D. PROP. XXV. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the base CB greater than the base EF; the angle BAC shall be greater than the angle EDF. For, if it be not greater, it must either be equal to it, or less than it; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal* to EF; B * 4. 1. |