of the side BC, let the two straight lines BD, CD be drawn B A * 20.1. E + 4 Ax. C of a triangle * are greater than the third side, 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. PROPOSITION XΧΙΙ. † 20. 1. † 4 Ax. * 16, 1. PROB. To make a triangle of which the sides shall be equal See N. 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 Band C greater 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 20. 1. K at the point D, but unlimited towards F equal to B, and GH equal to C: from the centre F, at the distance FD, describe* the circle DKL; and from the BC * 3 Post. centre G, 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. * 15 Def. + Constr. +1 Ax. * 15 Def. † Constr. 22. 1. * 8. 1. See N. Because the point F is the centre of the circle DKL, FD is equal * to FK; but FD is equal to the straight line A; therefore FK is equal to A: again, because G is the centre of the circle LKH, GH is equal* to GK; 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 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. PROPOSITION XXIII. PROB. At a 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 angle; it is required to make an angle at the given point A, in the DCE. D B G In CD, CE, take any points D, E, and join DE; and make the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC; so that CD be equal to AF, CE to AG, and DE to FG: the angle FAG shall be equal to the angle DCE. Because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG, the angle DCE is equal * to the angle FAG. Therefore, 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. PROPOSITION 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 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. Because AB is equal + to DE, and AC + to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the 23. 1. 3.1. Hyp. + Constr. + Constr. the angle EGF; therefore the angle DFG is greater than EGF; therefore, 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 angle is 19. 1. subtended by the greater side, therefore the side ĘG is greater than the side EF: but EG was proved to be equal to BC; therefore BC is greater than EF. Therefore, if two triangles, &c. Q-E. D. PROPOSITION XXV. THEOR. If two triangles have two sides of the one, equal to See N. 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 BC greater than the base EF: the A angle BAC shall be greater than the angle EDF. D E F For if it be not greater, it must either be equal to it, or less than it: B but the angle BAC is not equal to the angle EDF, because then the base BC would be equal * to EF; but it is not; therefore the angle BAC is not equal to the angle EDF : neither † Hyp. 4.1. * 24.1. † Hyp. is it less, because then, the base BC would be less * than the base EF; but it ist not; therefore the angle BAC is not less than the angle EDF: and it was shewn, 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. PROPOSITION XXVI. THEOR. If two triangles have two angles of the one, equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each, then shall the other sides 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, each to each, viz. ABC to DEF, and BCA to EFD; also one side equal to one side: and first, let those sides be † 3. 1. † Hyp. † Hyp. * 4.1. + 1 Ax. + Hyp. + Hyp. * 4. 1. For if AB be not equal to DE, one of them must be greater than the other: let AB be the greater of the two, and make BG equal + to DE, and join GC: therefore, because BG is equal to DE, and BC † to EF, the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal + to the angle DEF; therefore the base GC is equal * to the base DF, and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle GCB is equal to the angle DFE: but DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG is equal to the angle BCA, the less to the greater; which is impossible; 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 equalt to the angle DEF; therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF; and also the third. angle BAC to the third angle EDF. 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 because BH is equal to EF, and AB A D † 8. 1. B HC E F tot DE, the two AB, BH are equal to the two DE, EF, † Hyp. Q. E. D. PROPOSITION XXVII. C th I month 18. THEOR. If a straight line, falling upon two other straight lines, make the alternate angles equal to one another, these two straight lines shall be parallel. Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another: AB shall be parallel to CD. being produced, will meet either towards For if it be not parallel, AB and CD, B, D, or towards A, C: let them be pro- C/E duced and meet towards B, D, in the point 15 G D. G; therefore GEF is a triangle, and its exterior angle AEF is being produced, do not meet towards B, D. In like manner |