Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Hong Kong

The altitudes $AD$ and $BE$ of acute triangle $ABC$ intersect at $H$ . Let $F$ be the intersection of $AB$ and a line that is parallel to the side $BC$ and goes through the circumcentre of $ABC$ . Let $M$ be the midpoint of $AH$ . Prove that $\angle CMF = 90^\circ$ .

Solution

Let $P$ and $N$ be the projection of $F$ and $O$ on $BC$ respectively. Recall that $ON = \frac{1}{2} AH$ . Therefore, $FP = AM = MH$ .
Since $FP$ and $AM$ are perpendicular to $BC$ , they are parallel. Thus, $AFPM$ is a parallelogram. This implies $MP \parallel AF$ , and hence $MP \perp CH$ . Also, we have $MH \perp PC$ . Therefore, $H$ is the orthocentre of $\triangle MPC$ . It follows that $PH \perp CM$ .
Now, as $FP \parallel MH$ and $FP = MH$ , we know that $FPHM$ is another parallelogram. This shows $FM \parallel PH$ , and hence $FM \perp CM$ . In other words,
$\angle FMC = 90^\circ$ .

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