We first consider $\triangle ABC$ 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
$$MA \cdot MC = ME^2.$$
That means $ME^2$ is equal to the power of $M$ with respect to the circumscribed circle of $\triangle ABC$. Further, since $ME$ is the length of the tangent from $M$ to the inscribed circle of $\triangle ABC$, so $ME^2$ 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 $\triangle ABC$.
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.