70th 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 $AB\ne AC$, 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., $B{M}^2=DM\cdot MP$, since $DB=BM+DM$, $DC=CM-DM=BM-DM$ and $DP=MP-DM$. Alternatively, but equivalently, $DP\cdot MP=M{P}^2-B{M}^2$, 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, $DP\cdot MP=PB\cdot PC=(BM+MP)(MP-CM)=(MP+BM)(MP-BM)=M{P}^2-B{M}^2$, as desired.