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Mongolian Mathematical Olympiad Round 3

Geometry Difficulty 5.5 AIME, harder Prove it Mongolia

Let $ABC$ be a triangle. Let $D$ be a point on $AB$ and let $E$ be a point on $AC$ such that $BCED$ is cyclic. Let $F$ be the intersection of the segment $BE$ and the circumcircle of $ADC$ and let $G$ be the intersection of $CD$ and the circumcircle of $ABE$ . Suppose that the segments $BG$ and $CF$ meets at $S$ . Show that $\angle FAS = \angle GAS$ .

(Proposed by Khulan Tumenbayar)

Solution

Since the quadrilaterals $ADFC$ , $DBCE$ , $AEGB$ are cyclic, we have
$\angle ADC = \angle AFC = \angle ADC = \angle AEB = \angle AGB.$
Because $\angle AEF = \angle AFC$ , we have $\triangle AEF \sim \triangle ACF$ . Hence $AF^2 = AE \cdot AC$ . In the same way, we have $AG^2 = AD \cdot AB$ . It follows that $DBCE$ is cyclic. Thus $AF = AG$ . Since $AF = AG$ , $\angle AFS = \angle AGS$ and $AS$ is common, we have $\triangle AFS = \triangle AGS$ . We conclude that $\angle FAS = \angle GAS$ .

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