Considering an acute triangle $ABC$, $E,F$ are the feet of the altitudes onto $BC$ and $AC$. $M$,$N$ are the midpoints of $BE$, $AF$. $p$ is perpendicular to $AC$ and passes through $M$. $q$ is perpendicular to $BC$ and passes through $N$. $P$ is the intersection of $p,q$. The task is to prove that $P$ is the midpoint of $EF$.
I've been playing around with this problem for a while now. The result feels very intuitive and symmetric but all I was able to do is prove some things about $p,q$ in relation to the altitudes. This is not of much help as I need to connect them to the points $E$ and $F$. I would really appreciate your help.
Connect $EF$. Let the intersection of $p$ and $EF$ be $P_1$, and the intersection of $q$ and $EF$ be $P_2$.
Then $P_1$ is the midpoint of $EF$, as $MP_1$ is a midsegment of triangle $EBF$. For the same reason, $P_2$ is the midpoint of $EF$. Therefore, $P_1,P_2$ coincide at the midpoint of $EF$, let's call it $P_0$.
And since $p,q$ can have at most one intersection, $P_0$ is indeed the intersection of $p,q$, that is, $P$.