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Problem 2301

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.7 Prove it IMO Hk TST · Hong Kong

$ABCD$ is a cyclic quadrilateral with $BC = CD$ . The diagonals $AC$ and $BD$ intersect at $E$ . Let $X$ , $Y$ , $Z$ and $W$ be the incentres of triangles $riangle ABE$ , $riangle ADE$ , $riangle ABC$ and $riangle ADC$ respectively. Show that $X$ , $Y$ , $Z$ and $W$ are concyclic if and only if $AB = AD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since $\angle BAE = \angle CAD$ (in view of $BC = CD$ ) and $\angle EBA = \angle DCA$ , we have $\triangle ABE \sim \triangle ACD$ . Since $X$ and $W$ are incentres of $\triangle ABE$ and $\triangle ACD$ respectively, they are corresponding points under this similarity. It follows that
$\frac{AX}{AW} = \frac{AB}{AC}.$
Similarly, we have $\frac{AY}{AZ} = \frac{AD}{AC}$ . Now,
$\begin{align*} & W, X, Y, Z \text{ are concyclic} \\ \Leftrightarrow \quad & AX \times AZ = AY \times AW \\ \Leftrightarrow \quad & \frac{AX}{AW} = \frac{AY}{AZ} \\ \Leftrightarrow \quad & \frac{AB}{AC} = \frac{AD}{AC} \\ \Leftrightarrow \quad & AB = AD. \end{align*}$
This completes the proof.

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