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 ; from B, C, the ends of the side BC, draw BD, CD, from the point D within the A; then BD+DC<BA+AC, and BDC> <BAC. A E B Prop. 19. In ABE, Prop. 19. In CED, Prod. BD to E. BA+AE>BE, add EC to each; BA + AC>BE+EC. add DB to each; EC+ BE>CD + DB. But it has been proved BA + AC> BE + EC, much more then is BA + AC > CD + DB. Prop. 15. Again, in ▲ CDE, ext. / BDC > int. ▲ CED, and in ABE, ext. / CED>int, / BAC, much more then is BDC > BAC, Therefore, if from the ends of, &c. Prop. 3. PROP. XXI. PROB. 22. 1 Eu. 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. (19 Prop.) Let A, B, C, be the 3 given straight lines. It is required to make a A, of which the sides shall be equal to A, B, C, respectively. From D take DE unlimited towards E. GHC. Post. 3. From cent. F, and dist. FD, descr. © DKL. From cent. G, and dist. GH, descr. O HLK. Join KF and KG. Then sides of ▲ KFG = A, B, C. sides of KFG are made = A, B, C, which was to be done. PROP. XXII. PROB. 23. 1 Eu. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. L Let AB be the given str. line, A a given pt. in it, DCE the given ; to make an at A in AB / DCE. In CD, CE, take any points D, E. Join DE. Constr. Prop. 7. A D CD, CE, DE = AF, AG, FG, ea. to ea. Therefore, at the given point A in AB, the angle FAG is made given ▲ DCE, which was to be done. PROP. XXIII. THEOR. 24. 1 Eu. 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 As, having the sides AB, AC = sides DE, DF, ea. to ea., but ▲ BAC >▲ EDF: then, base BC > base EF. Of the sides DE, DF, let DE ✈ DF, at D in the str. line DE, S AB, AC= ED, DG, ea. to ea. Hyp Constr. BC EG, Prop. 4. Ax. 9. Prop. 18. BC, BC > EF. PROP. XXIV. THEOR. 25. 1 Eu. 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 |