make DF equal to A, FG equal to B, and GH equal to C; (1. 3) and from the centre F, at the distance FD, describe the circle DKL; (post. 3) and from the centre G, at the distance GH, describe the circle HLK; and join KF, KG. Then the triangle KFG shall have its sides equal to the three straight lines A, B, C. DEMONSTRATION. Because the point F is the centre of the circle DKL, therefore FD is equal to FK; (def. 15) but FD is equal to the straight line A; (constr.) therefore FK is equal to A. (ax. 1) Again, because G is the centre of the circle LKH, therefore GH is equal to GK; (def. 15) but GH is equal to C; (constr.) therefore GK is equal to C; and FG is equal to B; (constr.) therefore the three straight lines KF, FG, GK, are equal to the three A, B, C. And therefore the triangle KFG has its sides KF, FG, GK, equal to the given straight lines A, B, C. Q. E. F. PROP. XXIII.-PROBLEM. At a given point in a given straight line to construct a rectilineal angle equal to a given rectilineal angle. (References-Prop. I. 8, 22.) 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 an angle at the point A in the straight line AB, equal to the rectilineal angle DCE. CONSTRUCTION Take in CD, CE, any points D, E, and join DE; make the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG. (1. 22.) Then the angle FAG shall be equal to the angle DCE. DEMONSTRATION Because DC, CE, are equal to FA, AG, each to each, and the base DE to the base FG; (constr.) therefore the angle DCE is equal to the angle FAG. (1. 8.) Wherefore, at the given point A in the straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q.E. F. PROP. 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 the two sides of one of them greater than the angle contained by the two sides, equal to them, of the other; then the base of that which has the greater angle, shall be greater than the base of the other. (References-Prop. 1. 3, 4, 5, 19, 23; ax. 9.) Let ABC, DEF, be two triangles which have the two sides AB, AC, equal to the two DE, DF, each to each-viz., 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. 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) and make DG equal to AC or DF, (1. 3) and join EG, GF. DEMONSTRATION 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; (constr.) therefore the base BC is equal to the base EG. (1. 4.) And, because DG is equal to DF, therefore 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 the angle 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 EG has been proved to be equal to BC; therefore BC is greater than EF. Therefore, if two triangles, &c. Q. E. D. PROP. XXV.-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 ; then 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. (References-Prop. I. 4, 24.) 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 equal to DE, and AC to DF; but the base CB greater than the base EF. Then the angle BAC shall be greater than the angle EDF. DEMONSTRATION For if the angle BAC be not greater than the angle DEF, it must either be equal to it or less. Now, if the angle BAC be equal to the angle EDF, then must the base BC be equal to the base EF; (1. 4) but it is not; (hyp.) therefore the angle BAC is not equal to the angle EDF. Again, if the angle BAC be less than the angle EDF, then must the base BC be less than the base EF; (1. 24) but it is not; (hyp.) therefore the angle BAC is not less than the angle EDF; and it has been proved 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.— THEOREM. 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 side adjacent to the equal angles in each, or the side opposite to them; then the other sides shall be equal, each to each, and also the third angle of the one equal to the third angle of the other. (References - Prop. I. 3, 4, 16; ax. 1.) Let the two triangles ABC, DEF, have the angles ABC, BCA, equal to the angles DEF, EFD, each to each; also one side equal to one side. First. Let the sides adjacent to the equal angles in each be equal; viz., BC equal to EF. Then shall the side AB be equal to the side DE, the side AC to the side DF, and the angle BAC to the angle EDF. A A B CONSTRUCTION For if AB be not equal to DE, one of them is greater than the other. Let AB be the greater of the two, and make BG equal to DE, (1. 3) and join GC. DEMONSTRATION Then in the two triangles GBC, DEF, because BG is assumed to be equal to DE, and BC is equal to EF, (hyp.) the two sides GB, BC, must be equal to the two DE, EF, each to each; and the angle GBC is equal to the angle DEF; (hyp.) therefore the base GC must be equal to the base DF, and the triangle GBC to the triangle DEF, and the remaining angles of the one equal to the remaining angles of the other, each to each, to which the equal sides are opposite; therefore the angle GCB must be equal to the angle DFE; (1. 4) but the angle DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG must be equal to the angle BCA, (ax. 1) the less to the greater, which is impossible; therefore AB is not unequal to DE, that is, AB is equal to DE. Therefore in the triangles ABC, DEF, because the two AB, BC, are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; (hyp.) therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. (1. 4.) Next, let the sides which are opposite to the equal angles in each triangle be equal to one another, viz., AB to DE. Then likewise in this case the other sides shall be equal, AC to DF, and BC to EF; and also the third angle BAC to the third angle EDF. |