National Competition

Question

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.

Accepted Answer

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