Hyp. Ax. 3. but ABC + ▲ ABD = 2 rt. ▲ ; ABC + ABE = 2 ABC+ABD. From these equals take away ABE: = or, less= ABC, ABD; greater, which is absurd; .. BE is not in the same str. line with BC. In like manner it may be shown, that no other line but BD can be in the same str. line with BC. Wherefore, if at a point, &c. PROP. XIV. THEOR. 15. 1 Eu. If two straight lines cut one another, the vertical or opposite angles shall be equal. Let the str. lines AB, CD, cut one another in E; then AEC = / DEB, and ▲ CEB = AED. ..str.line AE makeswith CDthes CEA,AED, Prop. 12... CEA + 2 AED 2 rt. S. Again, ...str.lineDEmakes with ABthes AED,DEB. Prop. 12... AED + ▲ DEB = 2 rt. s, Ax. 1. :: Z CEA + 2 AED = 2 AED+ 2 DEB. From these equals take away the common .. remain. CEA = remain. ▲ DEB. In the same manner it may be shown, that Z CEB AED. Therefore, if two straight lines, &c. COR. 1.-If two str. lines cut one another, thes which they make at the pt. where they cut, are together equal to 4 rt. s. COR. 2.-All the angles made by any number of lines meeting in one pt., are together equal to 4 rt. s. Ax. 3. PROP. XV. THEOR. 16. 1 Eu. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a A, and the side BC be prod. to D; then the ext. ▲ ACD > ▲ CBA or BAC, the int. oppo. A s. F Prop. 14. but ACD > < ECF; < ACD > < BAC. In the same way, if BC be bisected, and AC be prod. to G, it may be shown, that LBCG or ACD > ▲ ABC. Therefore, if one side, &c. PROP. XVI. THEOR. 17. 1 Eu. Any two angles of a triangle are together less than two right angles. Let ABC be any A, any two of its s are together less than 2 rt. s. In like manner it may be proved, that and CAB+ ▲ ABC <2 rt. s. PROP. XVII. THEOR. 18.1 Eu. The greater side of every triangle is opposite to, or subtends the greater angle. PROP. XVIII. THEOR. 19. 1Eu. The greater angle of every triangle is subtended by the greater side; or has the greater side opposite to it. Prop. 5. By Hyp. Prop. 17. By Hyp. If ABC be a▲, of which ABC > < BCA; then AC>AB. For AC must be either >,=, or < AB. then ABC = ACB. Any two sides of a triangle are together greater than the third side. Let ABC be a ▲, then AB + AC > BC, AB+ BC > AC, D B |