HMMT November

Geometry Difficulty 4.7 AIME Find the answer

Let $A B C D$ be a parallelogram with $A B=480, A D=200$ , and $B D=625$ . The angle bisector of $\angle B A D$ meets side $C D$ at point $E$ . Find $C E$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

First, it is known that $\angle B A D+\angle C D A=180^{\circ}$ . Further, $\angle D A E=\frac{\angle B A D}{2}$ . Thus, as the angles in triangle $A D E$ sum to $180^{\circ}$ , this means $\angle D E A=\frac{\angle B A D}{2}=\angle D A E$ . Therefore, $D A E$ is isosceles, making $D E=200$ and $C E=280$ .

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