China Mathematical Olympiad

We first consider $△ABC$ and segment $PFD$. By Menelaus theorem we have $$\frac{CP}{PA}\cdot \frac{AF}{FB}\cdot \frac{BD}{DC}=1.$$ Then $$PA=CP\cdot \frac{AF}{FB}\cdot \frac{BD}{DC}=(PA+b)\frac{p-a}{p-c}.$$ (We define $a=BC$, $b=CA$, $c=AB$, $p=\frac{1}{2}(a+b+c)$; and without loss of generality, assume $a>c$.) Then we get $$PA=\frac{b(p-a)}{a-c}.$$ Further, $$PE=PA+AE=\frac{b(p-a)}{a-c}+p-a=\frac{2(p-c)(p-a)}{a-c},$$ $$ME=\frac{1}{2}PE=\frac{(p-c)(p-a)}{a-c},$$ $$MA=ME-AE=\frac{(p-c)(p-a)}{a-c}-(p-a)=\frac{(p-a{)}^2}{a-c},$$ $$MC=ME+EC=\frac{(p-c)(p-a)}{a-c}+(p-c)=\frac{(p-c{)}^2}{a-c}.$$ Then we have $$MA\cdot MC=M{E}^2.$$ That means $M{E}^2$ is equal to the power of $M$ with respect to the circumscribed circle of $△ABC$. Further, since $ME$ is the length of the tangent from $M$ to the inscribed circle of $△ABC$, so $M{E}^2$ is also the power of $M$ with respect to the inscribed circle. Hence, $M$ is on the radical axis of the circumscribed and inscribed circles of $△ABC$. In the same way, $N$ is also on the radical axis. Since the radical axis is perpendicular to $OI$, then $OI\perp MN$. That completes the proof.