1997 OIM Problems/Problem 5

Problem

In an acute triangle $ABC$, let $AE$ and $BF$ be two heights, and let $H$ be the orthocenter. The symmetrical line of $AE$ with respect to the (interior) bisector of the angle at $A$ and the symmetrical line of $BF$ with respect to the (interior) bisector of the angle at $B$ intersect at a point $O$. The lines $AE$ and $AO$ intersect a second time the circumference circumscribes triangle $ABC$ at points $M$ and $N$, respectively.

Let $P$ be the intersection of $BC$ with $HN$; $R$, the intersection of $BC$ with $OM$; and $S$ the intersection of $HR$ with $OP$.

Prove that $AHSO$ is a parallelogram.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

https://www.oma.org.ar/enunciados/ibe12.htm