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Geometry Difficulty 4.8 AIME Prove it Japan

In isosceles right triangle $ABC$ with $\angle BAC = 90^\circ$ and $AB = AC = 7$ , let $D$ , $E$ and $F$ be points on sides $BC$ , $CA$ and $AB$ respectively. Given that $\angle EDF = 90^\circ$ , $DE = 5$ , $DF = 3$ , find the length of $BD$ .

Solution

Let $P$ be a point on line $AB$ such that $PD \perp BC$ , then we have $\angle CDE = \angle PDF$ since $\angle CDP = \angle EDF = 90^\circ$ , and $\angle DCE = \angle DPF = 45^\circ$ , which yields $\triangle PDF \sim \triangle CDE$ with a similarity ratio of $3:5$ . In addition, we have $BD = DP$ since $\angle PBD = \angle BPD = 45^\circ$ . Therefore it follows $BD : DP : DC = 3 : 3 : 5$ , and so we have $BD = 7\sqrt{2} \cdot \frac{3}{8} = \frac{21\sqrt{2}}{8}$ .

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