$AD$, $BE$, $CF$ are the altitudes of $\triangle ABC$. If $P$, $Q$, $R$ are the mid-points of $DE$, $EF$, $FD$, respectively, then show that the perpendicular from $P$, $Q$, $R$ to $AB$, $BC$, $CA$, respectively, are concurrent.
I first thought of Carnot's theorem, but I don't see anything about the lengths required.
So, I looked at $\triangle DEF$, I know $AD$, $BE$ and $CF$ are angle bisectors of the orthic triangle. Therefore, the problem reduces to proving, lines parallel to angle bisectors, passing through midpoints of the respective sides are concurrent.
To solve this, I thought of using a phantom point. Let lines through $Q$ and $R$, parallel to $AD$ and $BE$ respectively intersect at $M$, let line through $M$, parallel to $CF$ intersect $DE$ at $P'$, it is to be proved that $P'$ is the midpoint of $DE$.
I tried some basic angle chasing but could not proceed further than this.
Best Answer
Proceeding from your first idea, let's consider the labeled figure below and $\triangle P'RQ$.
It is very easy to note that:
$$\angle D_1=\angle D_2=\angle Q_1, \\ \angle D_1=\angle Q_2.$$
Therefore, $QN$ is the angle bisector of $\angle RQP'$, and we are done because $M$ must be the incenter of $\triangle P'RQ$.