But we have also seen that the point B coincides with the point E; therefore the whole base BC shall coincide with the whole base EF. For, if the point B coincides with the point E, and the point C coincides with the point F, then, if the whole base BC does not coincide with the whole base EF, we have two straight lines enclosing a space, which is impossible (ax. 10). Therefore, 1. The base BC coincides with the base EF. 2. The triangle ABC coincides with the triangle DEF. And therefore, it is proved (ax. 8), as required, that 2. The triangle ABC: the triangle DEF. If two triangles, &c. Q. E. D. Exercises. 1. Given the straight lines AB and CD, of which AB is the greater; it is required to produce CD to make it = AB. 2. Prove Prop. IV. when the triangle DEF is applied to the triangle ABC. 6 PROP. V. THEOREM. The angles at the base of an isosceles triangle are equal to each other, and, if the equal sides are produced, the angles upon the other side of the base are also equal to each other. Let ABC be an isosceles triangle with the side AB= the side AC, and let AB and AC be produced to D and E respectively. Then it is to be proved that 1. The angles ABC and ACB, at the base other. each 2. The angles DBC and ECB, upon the other side of the base, also each other. CONSTRUCTION.-1. In BD take any point F. 2. From AE cut off AG=AF (I. 3), and join BG and CF. PROOF.-Because in the triangles FAC and GAB we have the sides FA and AC, and their angle FAC, in the former = the sides GA and AB, and their angle GAB, in the latter, each to each (hyp. and cons.), therefore the base CF = the base BG, the angle ACF = the angle ABG, and the angle AFC the angle AGB (I. 4), i.e. the angle BFC = the angle CGB (note 2 def. 15). Next, because in the triangles BCF and CBG we have the sides BF and FC, and their angle BFC, in the former = the sides CG and GB, and their angle CGB, in the latter, each to each (cons. and proof above), therefore the angle BCF = the angle CBG, and the angle FBC = the angle GCB (I. 4), i.e. the angle DBC = the angle ECB (note 2 def. 15), and these are the angles upon the other side of the base. Further, because the angle ABG = the angle ACF, and the angle CBG = the angle BCF, as already proved, therefore if the angle CBG be taken from the angle ABG, and the angle BCF be taken from the angle ACF, then the remaining angle ABC = the remaining angle ACB (ax. 3), and these are the angles at the base. Therefore, it is proved, as required, that 1. The angles ABC and ACB, at the base each other, and 2. The angles DBC and ECB, upon the other side of the base, also each other. Wherefore, The angles at the base of an isosceles triangle, &'c. Q. E. D. Cor. Every equilateral triangle is also equiangular. NOTE. It may assist the scholar, in learning this proposition, to observe that in the first step in the proof the larger pair of triangles, FAC and GAB, and in the second step the smaller pair, BCF and CBG, are taken. He will also notice that the equality of the angles 'on the other side of the base' is proved, before the equality of those at the base' is demonstrated. Exercise. Given an isosceles triangle BAC with the vertical angle at A bisected by AD drawn to BC; prove that AD is perpendicular to BC. PROP. VI. THEOREM. If two angles of a triangle are equal to each other, then the sides also which subtend, or are opposite to, the equal angles, are equal to each other. Let ABC be a triangle having the angle ABC = the angle ACB : Then it is to be proved that The side AB = the side AC. B D CONSTRUCTION. Suppose that AB is greater than AC. From AB cut off a part DB = AC (I. 3) and join CD. PROOF.-Because in the triangles DBC and ACB we have the sides DB and BC and their angle DBC, in the former = the sides AC and CB, and their angle ACB, in the latter, each to each (hyp. and cons.), therefore the triangle DBC the triangle ACB (I. 4), i.e. a part the whole, which is absurd (ax. 9). = = Therefore, the supposition that AB is greater than AC is absurd. Similarly the supposition that AB is less than AC might be shown to be absurd. Therefore, it is proved, as required, that Wherefore, The side AB the side AC. If two angles of a triangle, &c. Q. E. D. Cor.-Every equiangular triangle is also equilateral. N.B.-The proof here given is an instance of what is called Indirect Demonstration. This means, the truth of the statement asserted is only proved by showing that absurdity follows from an argument based on the supposition that such statement is not true. The scholar must be careful to notice that the term 'absurd,' which is found in this method of proof, does not apply to the entire demonstration, but only to the hypothesis on which it is based. The chain of argument beyond this, in these cases, is strictly correct. Exercises. 1. Prove the above Proposition on the supposition that AC is greater than AB. 2. If a straight line DA be drawn at right angles to another straight line BC from its middle point D, prove, if BA and CA be joined, that BA = CA. |