Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let $M$ be the midpoint of the side $BC$. If $AB = AC$, then $P$ is the ideal point of the line $BC$, the points $Q$ and $R$ fall at $C$ and $B$, respectively, and the circle $PQR$ degenerates into the line $BC$ on which $M$ clearly lies.

Assume henceforth that $AB \neq AC$, say, $AB > AC$. It is clearly sufficient to show that
$$DM \cdot DP = DQ \cdot DR.$$
Since $EF$ and $QR$ are parallel, and $B, C, E, F$ are concyclic ($E$ and $F$ both lie on the circle on diameter $BC$), so are $B, C, Q, R$. Hence $DB \cdot DC = DQ \cdot DR$, and it is therefore sufficient to show that $DB \cdot DC = DM \cdot DP$, i.e., $BM^2 = DM \cdot MP$, since $DB = BM + DM$, $DC = CM - DM = BM - DM$ and $DP = MP - DM$. Alternatively, but equivalently, $DP \cdot MP = MP^2 - BM^2$, since $DM = MP - DP$.

The points $D, E, F, M$ are concyclic (they all lie on the nine-point circle of the triangle $ABC$), so $PD \cdot PM = PE \cdot PF$. The points $B, C, E, F$ are also concyclic (recall that $E$ and $F$ both lie on the circle on diameter $BC$), so $PE \cdot PF = PB \cdot PC$. Consequently, $DP \cdot MP = PB \cdot PC = (BM + MP)(MP - CM) = (MP + BM)(MP - BM) = MP^2 - BM^2$, as desired.