$\triangle{ABC}$ is an acute triangle, and $R$ is the foot of the altitude from $C$. If $H$ and $K$ are the reflections of $R$ across $\overline{\mathit{BC}}$ and $\overline{\mathit{AC}}$, respectively, and if $P$ and $Q$ are the intersections of $\overline{\mathit{HK}}$ with $\overline{\mathit{BC}}$ and $\overline{\mathit{AC}}$, respectively, $P$ is the foot of the altitude from $A$, and $Q$ is the foot of the altitude from $B$.
I would appreciate an explanation to this statement. I know that $\overline{\mathit{CH}} \cong \overline{\mathit{CR}} \cong \overline{\mathit{CK}}$, and $\mathrm{m}\angle\mathit{HCK} = 2 \cdot \mathrm{m}\angle{C}$. (I doubt this is relevant - $\mathrm{m}\angle{A} = \mathrm{m}\angle\mathit{CAK}$ and $\mathrm{m}\angle{B} = \mathrm{m}\angle\mathit{CBH}$.)
Since $$\angle CRQ = \angle QKC = \angle QHC$$ we have cyclic quadrilateral $QRHC$. But $B$ is also on this circle since $HCRB$ is also cyclic (because $\angle CHB + \angle CRB = 90+90 = 180$). So $BQ\bot AC$.