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Berkeley Math Circle: Monthly Contest 2

Geometry Difficulty 4.5 AIME Prove it United States

Problem:

Let $ABCDE$ be a convex pentagon with $CD = DE$ and $\angle BCD = \angle DEA = 90^{\circ}$ . Point $F$ lies on $AB$ such that $\frac{AF}{AE} = \frac{BF}{BC}$ . Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$ .

Solution

Solution:

Let $\omega$ denote the circumcircle of $\triangle CDE$ and let $D_{1}$ be the point opposite to $D$ . Let $DA$ meet $\omega$ at $A_{1}$ and let $F' = A_{1}C \cap AB$ . If we let $\alpha = \angle DA_{1}C = \angle DD_{1}C = \angle ED_{1}D$ then
$\frac{AF'}{AE} = \frac{AF' \cdot AD_{1}}{AA_{1} \cdot AD} = \frac{\sin \alpha \cdot \sin \angle D_{1}DA}{\sin \angle A_{1}F'A \cdot \sin \alpha} = \frac{\sin \angle D_{1}CA}{\sin \angle CF'B} = \frac{F'B}{BC}$
Hence $F = F'$ . Now $\angle F'CE = \angle A_{1}CE = \angle A_{1}DE = \angle ADE$ .

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