Required. To prove angle A CD equal to angle BCD. A D B II. Given.-Side A B equal to side A C. Base A E equal to base A D. D B Required. To prove angle A B E equal to angle ACD. III. Given. The parallelogram ABCD (see page 73). A D Required. To prove angle B A D equal to angle ADC. THEOREM (Euclid I. 24). Repeat. The enunciations of Euc. I. 4, 5, and 19; and Axiom 9. General Enunciation. 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 the one, greater than the angle contained by the two sides of the other, the base of the one with the greater angle shall be greater than the base of the other. The same, in tabular form. GIVEN. Two triangles, having (a) two sides of one equal to two sides of the other, each to each ; (b) The angle contained by the two sides of one, greater than the angle contained by the two sides of the other. REQUIRED. To prove that— The base of the triangle with the greater angle is greater than the base of the other triangle. Particular Enunciation (see Fig. 1, next page). Given. The two triangles ABC, DEF, having (a) The side A B equal to the side DE, (b) The side A C equal to the side DF, and (c) The angle BAC greater than the angle EDF. Required. To prove that the base B C is greater than the base E F Construction. In the triangle DEF (Fig. 1) the two sides are DE and DF. Of those two sides one must be either equal to or greater than the other (see Exercise). Fig.1 ΔΔ B Let DE be the side which is not greater than the other. Then at the point D, in the straight line DE (Fig. 2), (d) make the angle EDG equal to the angle BAC (by Euc. I. 23). Fig. 2 NA B E F (e) Make D G equal to DF, and therefore equal to A C. Join EG and GF. Proof. A B is equal to DE (Given a). A C is equal to DG (Construction e). (ƒ) .·. The two sides B A, A C are equal to the two sides E D, D G. (g) And the included angle BAC is equal to the included angle E D G (Construction d). (h). (by Euc. I. 4) the base BC is equal to the base E G. Again (by Euc. I. 5) D G is equal to DF (Construction e). (i) ... angle D G F is equal to angle D F G. But angle E G F is a part of angle D G F. (k) .. angle DGF is greater than angle EGF (Axiom 9). (7) .. angle DFG is greater than angle EGF (see i). But, angle DFG is a part of angle E FG. (m). angle EFG is greater than angle DFG (Axiom 9). Now, from Conclusions / and m, angle E F G is greater than angle E G F. And (by Euc. I. 19) The greater angle of a triangle has the greater side opposite to it. ... base E G is greater than base E F. But E G is equal to B C (h). ... base B C is greater than base E F. Q. E. D. EXERCISES.-I. Draw the triangles A B C, DEF (Fig. 2), with A B, A C, DE, D F, all equal. II. In Fig. 2 draw a circle from centre D at radius DF, and then show that the line E G must fall above the point F. G THEOREM (Euclid I. 25). Repeat. The enunciations of Euc. I. 4 and 24. General Enunciation. 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. The same, in tabular form. GIVEN. Two triangles, having (a) two sides of one equal to two sides of the other, each to each; (b) but the base of one greater than the base of the other. Particular Enunciation. REQUIRED. To prove that— The angle contained by the two sides of the triangle with the greater base, is greater than the angle contained by the two sides of the other triangle. Given. The two triangles A B C, DE F, having (a) A B equal to DE, (b) A C equal to DF, (c) BC greater than EF Required. To prove that angle B A C is greater than angle EDF. Proof. The angle B A C must be either I. equal to angle EDF, or |