PROPOSITION 21.-Theorem. If from the ends of a 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 triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle. Then BD and DC shall be less than BA and AC, the other two sides of the triangle, but shall contain an angle BDC greater than the angle BAC. E B Construction. Produce BD to meet the side AC in E. Demonstration. Because two sides of a triangle are greater than the third side (I. 20), therefore 1. The two sides BA, AE of the triangle ABE are greater than BE; to each of these unequals add EC, therefore 2. The sides BA, AC are greater than BE, EC (Ax. 4). Again, because 3. The two sides CE, ED of the triangle CED are greater than DC (I. 20); add DB to each of these unequals; therefore 4. The sides CE, EB are greater than CD, DB (Ax. 4). But it has been shown that BA, AC are greater than BE, EC, much more then 5. BA, AC are greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle (I. 16), therefore for the same reason, 1. The exterior angle BDC of the triangle CDE is greater than the interior and opposite angle ČED; 2. The exterior angle CED of the triangle ABE is greater than the interior and opposite angle BAC; and it has been demonstrated that the angle BDC is greater than the angle CEB, much more therefore 3. The angle BDC is greater than the angle BAC. Therefore, if from the ends of the side, &c. Q.E.D. PROPOSITION 22.-Problem. 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. Let A, B, C be the three given straight lines, of which any two whatever are greater than the third (I. 20), namely, 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, B Construction. Take a straight line_DE terminated at the point D, but unlimited towards E, make DF equal to A, FG equal to B, and GH equal to C (I. 3); from the centre F, at the distance FD, describe the circle DKL (Post. 3); from the centre G, at the distance GH, describe the circle HLK; from K where the circles cut each other, draw KF, KG to the points F, G. Then the triangle KFG shall have its sides equal to the three straight lines A, B, C. Proof. Because the point F is the centre of the circle DKL, therefore 1. FD is equal to FK (Def. 15), but FD is equal to the straight line A, therefore 2. FK is equal to A. Again, because G is the centre of the circle HKL, therefore 3. GH is equal to GK (Def. 15), but GH is equal to C; therefore also 4. GK is equal to C (Ax. 1), A B с and FG is equal to B (constr.), therefore the three straight lines KF, FG, GK are respectively equal to the three A, B, C ; and therefore 5. The triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. Q.E.F. E PROPOSITION 23.-Problem. 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 ▲ the given point in it, and DCE the given rectilineal angle. B It is required, at the given point A in the given straight line AB, to make an angle that shall be equal to the given rectilineal angle DCE. Construction. In CD, CE take any points D, E, and join DE; on AB make the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so that AF be equal to CD, AG to CE, and FG to DE (I. 22). Then the angle FAG shall be equal to the angle DCE. Proof. Because FA, AG are equal to DC, CE, each to each, and the base FG is equal to the base DE, therefore 1. The angle FAG is equal to the angle DCE (I. 8). Wherefore at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q.E.F. PROPOSITION 24.-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 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. B Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, namely, AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF. Then the base BC shall be greater than the base EF. Construction. Of the two sides DE, DF, let DE be not greater than DF, at the point D in the straight line DE, and on the same side of it as DF, make the angle EDG equal to the angle BAC (I. 23), make DG equal to DF or AC (I. 3), and join EG, GF. C D Demonstration. Then, because DE is equal to AB, and DG to AC, the two sides DE, DG are equal to the two AB, AC, each to each, and the angle EDG is equal to the angle BAC (constr.); therefore B 1. The base EG is equal to the base BC (I. 4). And because DG is equal to DF in the triangle DFG, therefore 2. The angle DFG is equal to the angle DGF (I. 5) ; but the angle DGF is greater than the angle EGF (Ax. 9); therefore 3. The angle DFG is greater than the angle EGF; much more therefore 4. The angle EFG is greater than the angle EGF. And because in the triangle EFG, the angle EFG is greater than the angle EGF, and that the greater angle is subtended by the greater side (I. 19); therefore 5. The side EG is greater than the side EF; but EG was proved equal to BC; therefore 6. BC is greater than EF. Wherefore, if two triangles, &c. Q.E.D. PROPOSITION 25.-Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, the angle contained by the sides of the one 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, namely, AB equal to DE, and AC to DF; but the base BC greater than the base EF. |