Let $BE$ and $CF$ be the altitudes of acute triangle $ABC$. Let $H$ be the orthocenter of $ABC$ and $M$ be the midpoint of side $BC$. The points of intersection of the midperpendicular line to $BC$ with segments $BE$ and $CF$ are denoted by $K$ and $L$ respectively. The point $Q$ is the orthocenter of triangle $KLH$. Prove that $Q$ belongs to the median $AM$.
(Bohdan Zheliabovskyi)
Let $P$ denote the centre of spiral similarity $f$ sending segment $KL$ to $BC$. Note that $P$ lies on circles $(BKM){}$ and $(CLM){}$. Moreover, these two circles pass through $F, E$ respectively since \[\angle BFK = \angle CEL = \angle BMK = \angle BML = 90^{\circ}\]Now note that $\triangle HLK$ has angles equal to $\triangle ABC{}$. It follows that $\triangle HKL \sim \triangle ABC$, in particular, $f$ sends $\triangle HKL$ to $\triangle ABC{}$. The next fact kills the problem:
Fact (?) : Let $X$ be any point. Then $P, X, f^2(X)$ are collinear.
Proof: Well, the spiral sim rotates about $P$ by $90^{\circ}$ so this is trivial.
Now note that $\tau(Q) = H$ and $\tau(H) = A$. Thus $A = f^2(Q)$, implying that $A, P, Q$ are collinear. Therefore it suffices to show that $P$ lies on the median. This is clear by radax on $(BCEF), (CEPM), (BFPM)$. $\square$