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Romanian Mathematical Olympiad - District Round

Geometry Difficulty 4.9 AIME Prove it Romania

Let $ABCD$ be a square. Point $E$ lies in the interior of the angle $\angle CAB$ such that angle $\angle BAE$ is $15^\circ$ , and lines $BE$ and $BD$ are perpendicular. Show that $AE = BD$ .

Solution

Lines $AC$ and $BE$ are parallel, because both are perpendicular to $DB$ . Let $F$ be the foot of the perpendicular from $E$ onto $AC$ . Notice that $EF = BO = \frac{BD}{2}$ , where $O$ is the centre of the square. The triangle $FAE$ has a right angle at $F$ and has $\angle FAE$ of $30^\circ$ , hence $AE = 2EF = BD$ , as needed.

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