THEOREM (Euclid I. 5). Definition. An isosceles triangle has two sides equal. Repeat. The enunciation of Euc. I. 4, and Axiom 3. General Enunciation. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another. The same, in tabular form. GIVEN. (a) An isosceles triangle (b) having the equal sides produced. REQUIRED. To prove that— (a) The angles at the base are equal. (b) The angles on the other side of the base are equal. Particular Enunciation. Given. The isosceles triangle (a) A B C, having A B equal to AC, and the sides AB, AC produced to D and E (Fig. 1). Required. To prove that (b) angle A B C is equal to angle A CB, (c) angle D B C is equal to angle E CB. Construction. In BD take any point F(Fig. 2). (d) From A E cut off a part A G equal to AF (by Euc. I. 3). Join BG and FC. (The difficulty of this proposition arises from the fact that there are several triangles partly overlying each other. A B F Fig.1. E Thus we have to deal with two pairs of triangles. The two triangles ABG, ACF: and the two triangles BCF, CDG. You will understand this at once if you B C E cut out four triangles in pasteboard of the same shape as those in the figure, and lay them over one another as they are there.) Proof. Let us apply Euc. I. 4 to each of the pairs of triangles. First, we will take triangles ABG, ACF (In Fig. 3 those triangles are shown separately, as they would appear if lifted off one another.) Fig. 3. AA B G In these triangles we haveAF equal to AG (Construction d), AC equal to AB (Given a). angle FAC equal to angle GAB. (These angles are equal because, in Fig. 2, where they lie one on the other, they coincide.) .. by Euc. I. 4, we have (e) base FC equal to base B G. Now, we take triangles B CF, CB G (Fig. 2). (In Fig. 4 those triangles are shown separately, as they would appear if lifted off one another. A F is equal to A G (by Construction d). A B is equal to A C (Given a). (h). The remainder B F is equal to the remainder CG (Axiom 3). Then, in the triangles B CF, CBG we have BF equal to CG (h), and angle B FC equal to angle C G B (ƒ). A A by Euc. I. 4, (i) angle FBC is equal to angle GCB, and (k) angle FCB is equal to angle G B C. But it has been proved that angle ABG is equal to angle ACF (g). (7) .. angle ABC is equal to angle ACB (k and g). But, angles A B C, A C B (1) are the angles at the base, and they are equal. And angles FBC, GCB (i) are the angles on the other side of the base, and they are equal. Q. E. D. EXERCISE. The pupil should now be asked to go through the proof, referring only to Fig. 2. THEOREM (Euclid I. 6). Repeat.-The enunciation of Euc. I. 4 and Axiom 9. General Enunciation. If two angles of a triangle be equal to one another, the sides which are opposite to the equal angles shall be equal to one another. Particular Enunciation. Given. The triangle ABC having (a) the angle ABC equal to the angle AC B. с Required. To prove that the side A B is equal to the side A C Construction. Let us suppose that A B is not equal to A C; Let us suppose that A B is greater than A C. Join DC. B |