Slovenian Mathematical Olympiad, 2016

There are two different configurations of this problem depending on whether the point $P$ lies inside the triangle $ABC$ or outside, so we use directed angles in our solution.
Let $K_1$ denote the circle centered at $H$ with radius $HD$. Let $K_2$ denote the circle centered at $A$ with radius $AD$. Let $E$ and $F$ be the feet of the altitudes of the triangle $ABC$ from $B$ and $C$ respectively. We show that the points $P, R, E$ and $F$ are collinear.
From $\angle HFA = \angle HEA = \frac{\pi}{2}$ it follows that the points $A, F, H$ and $E$ are concyclic.
Also, $\angle AEB = \angle ADB = \frac{\pi}{2}$ implies that $A, B, D$ and $E$ are concyclic.
The tangent $BP$ is perpendicular to the radius $HP$ of the circle $K_1$, so $\angle BPH = \frac{\pi}{2}$. This implies that $\angle BPH = \angle BFH = \angle BDH = \frac{\pi}{2}$, so the points $B, D, H, P$ and $F$ are concyclic. Similarly, the tangent $CR$ is perpendicular to the radius $AR$ of the circle $K_2$, so $\angle ARC = \angle AFC = \angle ADC = \frac{\pi}{2}$ and the points $A, F, D, C$ and $R$ are concyclic as well.
The lengths of the tangents from their intersection to the point where they touch the circle are equal, so $|BP| = |BD|$, and the distance from the centre to points on the circle is constant, so $|HP| = |HD|$. The triangles $BPH$ and $BDH$ match in all three sides so they are congruent and $\angle HBP = \angle DBH$. From here and the concyclicities proven above it follows that
$$\angle HFP = \angle HBP = \angle DBH = \angle DBE = \angle DAE = \angle HAE = \angle HFE.$$
This implies that $E, F$ and $P$ are collinear.
Similarly, $|CR| = |CD|$ and $|AR| = |AD|$, so the triangles $ARC$ and $ADC$ are congruent and $\angle DAC = \angle CAR$. This implies that
$$\angle HFR = \angle CFR = \angle CAR = \angle DAC = \angle HAE = \angle HFE.$$
and the points $E, F$ and $R$ are collinear.
We have shown that the line $EF$ passes through the points $P$ and $R$, so $E, F, P$ and $R$ are collinear.