Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Croatia

Circles $k_1$ and $k_2$ intersect in points $A$ and $B$ . Line $l$ intersects circle $k_1$ in points $C$ and $E$ , and circle $k_2$ in points $D$ and $F$ in such a way that $D$ is between $C$ and $E$ , and $E$ is between $D$ and $F$ . Lines $CA$ and $BF$ intersect in point $G$ , and lines $DA$ and $BE$ in point $H$ .
Prove that $CF \parallel HG$ .

Solution

It suffices to show that $\angle ECA = \angle HGA$ .
Since the quadrilateral $ACBE$ is cyclic, we have $\angle ECA = \angle EBA$ , so it suffices to show that $ABGH$ is a cyclic quadrilateral.
From triangle $DEH$ we have $\angle DHE = 180^\circ - \angle EDH - \angle HED$ , i.e.
$\angle AHB = \angle DHE = 180^\circ - \angle FDA - (180^\circ - \angle CEB),$
so by using that $\angle CEB = \angle CAB$ (which holds because $ACBE$ is also cyclic) we get that
$\angle AHB = 180^\circ - \angle FDA - (180^\circ - \angle CAB) = 180^\circ - \angle FDA - \angle BAG.$

Quadrilateral ADBF is cyclic, so we have $\angle FDA = \angle FBA = \angle GBA$ . It follows that
$\angle AHB = 180^\circ - \angle FDA - \angle BAG = 180^\circ - \angle GBA - \angle BAG = \angle AGB.$
Therefore, ABGH is a cyclic quadrilateral, which finishes the proof.

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