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Taiwan IMO Selection Camp

Geometry Difficulty 4.5 AIME Prove it Taiwan

Let quadrilateral $ABCD$ be inscribed in a circle, and let its two diagonals $AC$ and $BD$ meet at point $E$ . Let ray $DA$ and ray $CB$ meet at point $F$ ; let $G$ be a point in the plane such that $ECGD$ is a parallelogram; let $H$ be the reflection of point $E$ with respect to line $AD$ . Prove that $D, H, F, G$ are concyclic.

Solution

We first prove that triangle $FDG$ is similar to triangle $FBE$ . Since $ABCD$ is cyclic, triangle $EAB$ is similar to triangle $EDC$ , and likewise $FAB$ is similar to $FCD$ . From the parallelogram $ECGD$ we get $GD = EC$ and $\angle CDG = \angle DCE$ . By the inscribed angle property, $\angle DCE = \angle DCA = \angle DBA$ . Therefore
$\angle FDG = \angle FDC + \angle CDG = \angle FBA + \angle ABD = \angle FBE, \\ \frac{GD}{EB} = \frac{CE}{EB} = \frac{CD}{AB} = \frac{FD}{FB}.$
Hence $\triangle FDG$ is similar to $\triangle FBE$ (SAS), and $\angle FGD = \angle FEB$ .

Since $H$ is the reflection of point $E$ with respect to line $FD$ , we obtain
$\angle FHD = \angle FED = 180^\circ - \angle FEB = 180^\circ - \angle FGD.$
From this it follows that $D, H, F, G$ are concyclic.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.