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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Let $ABCD$ be a parallelogram with $AB = 480$ , $AD = 200$ , and $BD = 625$ . The angle bisector of $\angle BAD$ meets side $CD$ at point $E$ . Find $CE$ .

Solution

Solution:

First, it is known that $\angle BAD + \angle CDA = 180^\circ$ . Further, $\angle DAE = \frac{\angle BAD}{2}$ . Thus, as the angles in triangle $ADE$ sum to $180^\circ$ , this means $\angle DEA = \frac{\angle BAD}{2} = \angle DAE$ . Therefore, $DAE$ is isosceles, making $DE = 200$ and $CE = 280$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.