IMO Hk TST

Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Hong Kong

Let $O$ and $H$ be respectively the circumcentre and orthocentre of $\triangle ABC$ . Let $A'$ , $B'$ and $C'$ be the midpoints of $BC$ , $CA$ and $AB$ respectively and $D$ , $E$ and $F$ be respectively the feet from $A$ , $B$ and $C$ to the opposite sides. Show that $OA' \cdot HD = OB' \cdot HE = OC' \cdot HF$ .

Solution

It is well-known that $OA' = \frac{1}{2} AH$ , etc. Therefore, we need to prove
$AH \times HD = BH \times HE = CH \times HF.$
Indeed, $AH \times HD = BH \times HE$ is true since $A$ , $B$ , $D$ , $E$ are concyclic. By symmetry, the other equality also holds.

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