PROPOSITION XXI. Theorem. 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. Steps of the Demonstration. 1. Prove that BA + AC > BE + EC, 2. that BE + EC > CD + DB, 3. that much more is BA + AC > CD + DB, 4. that L BDC > BAC. PROPOSITION XXII. D Problem. To make a triangle, the sides of which shall be equal to three given straight lines, but any two whatever of these must be greater than the third. 'A Steps of the Demonstration. 1. Prove that FK = A, 2. that GK = C, 3. that FG = B. PROPOSITION XXIII. Problem. At a given point in a given straight line to make a rectilineal angle equal to a given rectilineal angle. 44 Proved by Proposition VIII. PROPOSITION XXIV. Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by 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. A ES B Steps of the Demonstration. 1. Prove that (in As ABC, DEG) base bc = base EG, 2. that 2 DFG = DGF, 3 that Z DFG > Z EGF, 4. that much more Z EFG > EGF, 5. that :. Eg or bc > EF. PROPOSITION XXV. ( Argument ad absurdum). Theorem. 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. A D B Steps of the Argument. 1. Prove that L BAC cannot be = L EDF, 2. that < BAC cannot be < < EDF, 3. that . Z BAC must be > L EDF. PROPOSITION XXVI. ( Argument ad absurdum). Theorem. If two triangles have two angles of 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 opposite to the 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 B of the other. Steps of the Demonstration to Case 1st. Suppose that AB + DE, and that a part of AB, as BG = DE, { Then prove, on that supposition, , base gc = base DF, 1. that in As GBC, DEF A GBC = A DEF, land Z GCB = 2 DFE, 2. that Z BCG = Z BCA, i. e. less = greater; which shows the supposition to be false; and :. that AB = DE, 3. that in As ABC, DEF base ac = base df, and < BAC = L EDF. Suppose that BC + EF, and that a part of BC, as BH = EF, Then prove, on that supposition, (base an = base DF, 1. that in AS ABH, DEF A ABH = A DEF, land Z BHA = _ EFD. 2. that < BHA = _ BCA; i. e. the ex. Z BHA (of A AHC) = inter. and opp. < BCA; which shows the supposition to be false, and :: that BC = EF, 3. prove that ac = DF, and Z BAC = LEDF. PROPOSITION XXVII. ( Argument ad absurdum). Theorem. If a straight line falling upon two other E B G F D straight lines makes the alternate angles equal to one А another, these two straight lines shall be parallel. Steps of the Demonstration. Having made the supposition that AB # cd, and that they :: meet as in G, Prove, on that supposition, 1. that in the A EGF the exterior 2 AEF = inter. and opp. Z EFD, which is impossible, and :: the supposition is false. A and c, E B PROPOSITION XXVIII. Theorem. If a straight line falling upon two other straight lines, makes the exterior angle equal to the interior and opposite angle on the same side of the line, or makes the interior angles on the same side together equal to two right angles, the two D straight lines shall be parallel to one another. To prove the 1st part of the proposition, show by Proposition XV. that the alternate Zs are equal. Steps of the Demonstration to part 2nd. 1. Prove that ZS AGH + BGH = _S BGH + GHD, 2. that alternate _ AGH = alternate / GHD, and :. that AB || CD. H 3. |