IMO Hk TST

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

Let $\triangle ABC$ be a triangle. $M$ is the midpoint of $AC$ , $D$ is a point on $AB$ . $BM$ and $CD$ meet at $O$ , with $AB = CO$ . $E$ is a point on $AC$ such that $DE // BM$ . Prove that $AB \perp BC$ if and only if $ADOM$ is a cyclic quadrilateral.

Solution

Using the parallel lines, we have $\triangle ADE \sim \triangle ABM$ and $\triangle CMO \sim \triangle CED$ . This implies
$\frac{AB}{BD} = \frac{AM}{ME} = \frac{CM}{ME} = \frac{CO}{OD}.$
As $AB = CO$ , we obtain $BD = OD$ .
Now,
$\begin{align*} & \Leftrightarrow \quad \angle BAM = \angle DOB \\ & \Leftrightarrow \quad \angle BAM = \angle DBO \\ & \Leftrightarrow \quad \angle BAM = \angle ABM \\ & \Leftrightarrow \quad MB = MA. \end{align*}$
Since $MA = MC$ , this is equivalent to $\angle ABC = 90^\circ$ , i.e. $AB \perp BC$ as desired.

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