Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Hong Kong

In $\triangle ABC$ , let $AD$ be the angle bisector of $\angle BAC$ , with $D$ on $BC$ . The perpendicular from $B$ to $AD$ intersects the circumcircle of $\triangle ABD$ at $B$ and $E$ . Prove that $E, A$ and the circumcentre $O$ of $\triangle ABC$ are collinear.

Solution

By simple angle chasing, we find that
$\begin{align*} \angle BAE &= \angle BAD + \angle DAE = \frac{A}{2} + \angle DBE = \frac{A}{2} + B - \left(90^\circ - \frac{A}{2}\right) \\ &= A + B - 90^\circ = 90^\circ - C = \angle BAO. \end{align*}$
As $O$ and $E$ lie on the same side of $AB$ as $C$ in this case, this implies $A, O, E$ are collinear.

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