IMO Shortlist

Geometry Difficulty 8.4 Shortlist Prove it IMO

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$ . The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$ . Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$ . Prove that $D$ , $H$ , $F$ , $G$ are concyclic.

Solution

We show first that the triangles $FDG$ and $FBE$ are similar. Since $ABCD$ is cyclic, the triangles $EAB$ and $EDC$ are similar, as well as $FAB$ and $FCD$ . The parallelogram $ECGD$ yields $GD = EC$ and $\angle CDG = \angle DCE$ ; also $\angle DCE = \angle DCA = \angle DBA$ by inscribed angles. Therefore
$\begin{gathered} \angle FDG = \angle FDC + \angle CDG = \angle FBA + \angle ABD = \angle FBE, \\ \frac{GD}{EB} = \frac{CE}{EB} = \frac{CD}{AB} = \frac{FD}{FB} . \end{gathered}$
It follows that $FDG$ and $FBE$ are similar, and so $\angle FGD = \angle FEB$ .

Since $H$ is the reflection of $E$ with respect to $FD$ , we conclude that
$\angle FHD = \angle FED = 180^{\circ} - \angle FEB = 180^{\circ} - \angle FGD .$

This proves that $D$ , $H$ , $F$ , $G$ are concyclic.

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