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Problem 1795

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.3 Prove it Selection tests for the Balkan Mathematical Olympiad 2013 · Saudi Arabia · 2013

$ABCD$ is a cyclic quadrilateral and $\omega$ its circumcircle. The perpendicular line to $AC$ at $D$ intersects $AC$ at $E$ and $\omega$ at $F$ . Denote by $\ell$ the perpendicular line to $BC$ at $F$ . The perpendicular line to $\ell$ at $A$ intersects $\ell$ at $G$ and $\omega$ at $H$ . Line $GE$ intersects $FH$ at $I$ and $CD$ at $J$ . Prove that points $C$ , $F$ , $I$ , and $J$ are concyclic.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

To prove that $C$ , $F$ , $I$ , and $J$ are concyclic, it is equivalent to prove that $\angle CFI = \angle CJI$ or $\angle CFI + \angle CJI = 180^{\circ}$ , depending on the configuration. We will present here the proof for one configuration. The proof for the other configuration is similar.

Because $AFCH$ is cyclic, we have $\angle CFI = \angle CAH$ .

Because $\angle FEA = \angle FGA = 90^{\circ}$ , the quadrilateral $AFEG$ is cyclic, and therefore $\angle CAH = \angle EFG$ .

It remains to prove that $\angle EFG = \angle DJG$ , which is equivalent to proving that quadrilateral $DGFJ$ is cyclic.

But $\angle EGF = \angle EAF$ since $AFEG$ is cyclic. On the other hand, because $AFCD$ is cyclic, we deduce that $\angle EAF = \angle CDF$ . Therefore, $\angle EGF = \angle CDF$ , which proves that $DGFJ$ is cyclic.

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