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Brazilian Math Olympiad

Geometry Difficulty 4.7 AIME Prove it Brazil

Let $ABC$ be an acute triangle and $H$ is orthocenter. Let $D$ be the intersection of $BH$ and $AC$ and $E$ be the intersection of $CH$ and $AB$ . The circumcircle of $ADE$ meets the circumcircle of $ABC$ at $F \neq A$ . Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ concur at a point on line $BC$ .

Solution

By the angle bisector theorem, it suffices to prove that $\frac{BF}{FC} = \frac{BH}{HC}$ .
We have $\angle EFB = 180^\circ - \angle FEA = 180^\circ - \angle FDA = \angle FDC$ and $\angle FBE = \angle FBA = \angle FCA = \angle FCD$ , so triangles $BEF$ and $CDF$ are similar. Thus
$\frac{BF}{FC} = \frac{BE}{CD} = \frac{BH \cos \angle EBH}{CH \cos \angle DCH} = \frac{BH \cos(90^\circ - \angle BAC)}{CH \cos(90^\circ - \angle BAC)} = \frac{BH}{CH}$

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