National Competition

Geometry Difficulty 4.3 AIME Prove it Austria

We are given an acute triangle $ABC$ with $AB > AC$ and orthocenter $H$ . The point $E$ lies symmetric to $C$ with respect to the altitude $AH$ . Let $F$ be the intersection of the lines $EH$ and $AC$ .
Prove that the circumcenter of the triangle $AEF$ lies on the line $AB$ .

Solution

Let $\theta$ be the angle between $AF$ and the tangent $t$ at $A$ to the circumcircle of $AEF$ . By the inscribed angle theorem, we have $\angle FEA = \theta$ . Due to the reflection, we have $\angle ACH = \angle FEA = \theta$ . Because of $\angle ACH = \theta$ , the tangent $t$ is parallel to $CH$ and thus orthogonal to $AB$ . Therefore, the circumcenter of the triangle $AEF$ lies on $AB$ .

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