1999 CEMC Gauss (Grade 8) Problems/Problem 14

Problem

$ABC$ is an isosceles triangle in which $\angle A = 92^{\circ}$ . $CB$ is extended to a point $D$ . What is the size of $\angle ABD$ ?

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

$\text{ (A) }\ 88^{\circ} \qquad\text{ (B) }\ 44^{\circ} \qquad\text{ (C) }\ 92^{\circ} \qquad\text{ (D) }\ 136^{\circ} \qquad\text{ (E) }\ 158^{\circ}$

Solution 1

Let $x$ be $m\angle ABC$ . For the triangle to be isosceles, there has to be $2$ equal angles, so either $m\angle ABC = m\angle ACB$ , $m\angle ACB = 92^{\circ}$ , or $m\angle ABC = 92^{\circ}$ .

However, the last two are impossible because $92^{\circ} + 92^{\circ} = 184^{\circ}$ , which is more than the number of degrees in a triangle. Therefore, $m\angle ABC = m\angle ACB = x$ .