Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Romania

Let $ABC$ be a triangle in which $m(\hat{A}) = 135^\circ$ . The perpendicular to the line $AB$ erected at $A$ intersects the side $[BC]$ at $D$ , and the angle bisector of $\angle B$ intersects the side $[AC]$ at $E$ . Find the measure of $\widehat{BED}$ .

Traian Preda

Solution

Let $I \in (BE)$ such that $IA$ bisects the angle $\widehat{DAB}$ . We deduce that $ID$ is the bisector of the angle $\widehat{ADB}$ . A short computation shows that $m(\widehat{DIB}) = 135^\circ$ , hence triangles $ABE$ and $IBD$ are similar. It follows that $\frac{AB}{IB} = \frac{BE}{BD}$ , so that $\frac{AB}{EB} = \frac{BI}{BD}$ . Thus, triangles $ABI$ and $DBE$ are similar as well and, since $m(\widehat{BED}) = m(\widehat{BAI})$ , we infer that $m(\widehat{BED}) = 45^\circ$ .

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