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Belarusian Mathematical Olympiad

Geometry Difficulty 5.3 AIME, harder Prove it Belarus

Let $\triangle ABC$ be an acute triangle with the orthocenter $H$ . Let $D$ be the point such that $HABD$ is a parallelogram ( $AB \parallel HD$ , $AH \parallel BD$ ). Let $E$ be the point on the line $DH$ such that $AC$ bisects $HE$ . The line $AC$ meets the circumcircle of the triangle $DCE$ at $C$ and $F$ .
Prove that $EF = AH$ .

Solution

1. See IMO-2015 Shortlist, Problem G1.

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