Restated 2011 ToT wrote: In triangle $ABC$, points $D$, $E$ and $F$ are bases of altitudes from vertices $A$, $B$ and $C$ respectively. Points $P$ and $Q$ are the projections of $F$ to $AC$ and $BC$ respectively. Prove that the line $PQ$ bisects the segments $DF$ and $EF$. We claim that if $G=EF \cap PQ$, then $G$ is the midpoint of $EF$, in the same manner one can do same for $DF$. We have by trivial angle chase that $QFPC$ is cyclic and $\triangle FQC\sim\triangle EPF$. Thus, applying Law of Sines multiple times, we obtain \begin{align*} \frac{FG}{GE}&= \frac{\sin\angle FPG\cdot \frac{PF}{\sin \angle PGF}}{\sin\angle GPE\cdot \frac{PE}{\sin \angle PGE}}\\ &= \frac{\sin\angle FPG\cdot PF}{\sin\angle GPE\cdot PE}\\ &= \frac{\sin\angle FCQ\cdot PF}{\sin\angle QFC\cdot PE}\\ &= \frac{QF\cdot PF}{CQ\cdot PE}\\ &= 1. \end{align*}We are done. Remark. You can also only do angle chase, showing that $\angle FEP=\angle GPE$ and $\angle FPG=\angle GFP$, which would imply that $FG=GP=GE$.