Again, because the exterior angle of a triangle is greater than the interior and opposite angle (1. 16), the exterior angle BDC of the triangle CDE is greater than the angle CED: For the same reason, the exterior angle CEB of the triangle ABE is greater than the angle BAE: But it has been shewn that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. Wherefore, If from the ends &c. PROP. XXII. Q. E. D. 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 (1. 20). 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 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 straight line DE terminated at the point D, but unlimited towards E, and (1.3) make DF equal to A, FG equal to K B, and GH equal to C; from the D centre F, at the distance FD, de scribe the circle DKL, and from the centre G, at the distance GH, describe the circle HLK; and join KF, KG: the triangle KFG has 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 was made equal to A; therefore also FK is equal to A: Again, because G is the centre of the circle HLK, GH is equal to GK: but GH was made 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. Q. E. F. PROP. 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 at the given point A in the given straight line AB an angle, equal to the given rectilineal angle DCE. C Λ 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 (1. 22), so that AF shall be equal to CD, AG to CE, and FG to DE: the angle FAG shall be equal to the angle DCE. E For because DC, CE, are equal to FA, AG, each to each, and the base DE to the base FG-therefore (1.8) the angle DCE is equal to the angle FAG: Wherefore, at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE. Q. E. F. 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 the one 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 to DE, and AC to DF, but the angle BAC greater than the angle EDF: the base BC shall be also 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 (1. 23); make DG equal to AC or DF, and join EG, GF. F 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 angle BAC is equal to the angle EDG-therefore the base BC is equal to the base EG And because DF is equal to DG, the angle DFG is equal to the angle DGF (1. 5): But the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than 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 side is opposite to the greater angle (1. 19)-therefore the side EG is greater than the side EF: But BC was proved to be equal to EG; therefore also BC is greater than EF. Wherefore, 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. D 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 to DE, and AC to DF, but the base BC greater than the base EF: the angle BAC shall be greater than the angle EDF. A B D C E F For, if it be not greater, it must be either equal to it, or less: But the angle BAC is not equal to the angle EDF, because then the base BC would be equal to the base EF (1. 4); but it is nottherefore the angle BAC is not equal to the angle EDF: Neither is it less, because then the base BC would be less than the base EF (1. 24); but it is not-therefore the angle BAC is not less than the angle EDF: And it has been 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. PROP. 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 sides which are 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. A D Let ABC, DEF be two triangles, which have the angles ABC, ACB, equal to the angles DEF, DFE, each to each, viz. ABC to DEF, and ACB to DFE, and have also one side equal to one side, and first, let those sides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF: then the other sides shall be G B E F equal, each to each, viz. AB to DE, and AC to DF, and the third angle BAC to the third angle EDF. For, if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater, and make BG equal to DE, and join GC: Then, because BG is equal to DE, and BC to EF (Hyp.), 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 the angle DFE is equal to the angle ACB (Hyp.); therefore also the angle GCB is equal to the angle ACB, the less to the greaterwhich is absurd: 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-therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. D Next, let 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, viz. BC to EF, and AC to DF, and also the third angle BAC to the third angle EDF. B Н CE For if BC be not equal to EF, let BC be the greater, and make BH equal to EF, and join AH: Then because BH is equal to EF, and AB to DE (Hyp.), the two AB, BH are equal to the two DE, EF, each to each, and they contain equal angles (Hyp.)—there |