Let $M$ be the midpoint of the side $BC$. First, we want to show that $M$ is on the circumcircle of the triangle $DPQ$. For the intersection point $X$ of $PQ$ and $BC$, Menelaus theorem guarantees
$$\frac{BX}{XC} = \frac{CE}{EA} = \frac{AF}{FB} = 1.$$
Since $AF = AE$, $BD = BF$, $CD = CE$ and $BX \cdot CD = XC \cdot BD$, from which we get $2XB \cdot XC = (XB + XC) \cdot XD$ by replacing $BD$ and $CD$ with $XD - XB$ and $XC - XD$, respectively. It follows that $XB \cdot XC = XM \cdot XD$, as $\frac{XB+XC}{2} = XM$. Hence the point $M$ lies on the circumcircle of the triangle $DPQ$.

Let $O'$ be the midpoint of the line segment $O_1O_2$. It suffices to show that $O'$ is the circumcenter of the triangle $DPQ$. first, we will consider the case where $AB \neq AC$. In this case, we will prove it by showing that $O'$ is on both the perpendicular bisector of $PQ$ and the perpendicular bisector of $DM$. In fact, since $AI$ is perpendicular to both $PQ$ and $O_1O_2$, $PQ$ and $O_1O_2$ are parallel to each other. It follows that $O'$ is on the perpendicular bisector of $PQ$. Let $I_A, I_B$ and $I_C$ be the centers of excircles each of which is tangent to $BC$, $AC$ and $AB$, respectively. It is easy to see that the circumcircle of the triangle $ABC$ is the same as the nine point circle of the triangle $I_A I_B I_C$. Thus, if we let $O_3$ be the circumcenter of the triangle $IBC$ and let $S$ be the intersection point of $OO_3$ and $I_B I_C$, the point $S$ is the midpoint of the line segment $I_B I_C$. Note that the dilatation with center $I$ and ratio 2 sends the triangle $O_1O_2O_3$ to the triangle $I_C I_B I_A$, which implies that $O'$ is the midpoint of the line segment $IS$. Therefore we can conclude that $O'$ is on the perpendicular bisector of $DM$, as both $ID$ and $SM$ are perpendicular to $BC$.

Next, we will consider the case where $AB = AC$. Note that $O'$ is the midpoint of $AI$ in this case. Since $\angle AEI = 90^\circ = \angle AFI$, the points $A, F, I, E$ are on the circle with the center $O'$. Let $R$ be the radius of this circle. Then $O'D^2 - R^2 = DI \cdot DA = DA^2 - AI \cdot DA$ holds. The points $B, D, I, F$ being concyclic, we can see $AI \cdot DA = AF \cdot AB$. Since $DA^2 = AB^2 - BD^2$,
$$O'D^2 - R^2 = (AB^2 - BD^2) - AF \cdot AB = AB^2 - BF^2 - AF \cdot AB = BF \cdot AF.$$
Similarly, we can get $O'P^2 - R^2 = PE \cdot PF$. But $BF \cdot AF = QF \cdot PF = PE \cdot PF$, so
$$O'D^2 - R^2 = O'P^2 - R^2.$$
It easily follows that $O'D = O'P$. In a similar manner, we can show $O'D = O'Q$. Since $O'P = O'D = O'Q$, $O'$ is the circumcenter of the triangle $DPQ$.