Let $K=\odot(ABE)\cap\odot(ACD)$. An inversion around $A$, radius $\sqrt{AB\cdot AE}$ followed by a reflection across angle bisector of $\angle BAC$ sends $B\to E, C\to D$ and $K\to BE\cap CD$. Thus $K$ lies on $A$-symmedian of $\Delta ABC$. Moreover, since $AK\perp PQ$, rays $AK, AO$ are isogonal w.r.t. $\angle BAC$, implying the conclusion. BTW. Can someone post P1 and P3 please?

Let the second intersections of the line $DE$ with $\odot{(ABE)}$ and $\odot{(ACD)}$ be $X$ and $Y$ respectively, and let $R$ be the intersection of the tangents to $\odot{(ABC)}$ at $B,C$. Observe that $\measuredangle{ABX}=\measuredangle{AED}=\measuredangle{ACB}\implies R,B,X$ are collinear. Similarly the points $R,C,Y$ are also collinear. As $\measuredangle{RXY}=\measuredangle{BAC}=\measuredangle{XYR},$ we have $RX=RY$. Thus we have $$\text{pow}(R,\odot{(ABE)})=RB\cdot RX=RC\cdot RY=\text{pow}(R,\odot{(ACD)}),$$implying that $AR$ is the radical axis of $\odot{(ABE)}$ and $\odot{(ACD)}$. Hence we have $AR\perp O_1O_2\implies AR,AO$ are isogonal with respect to $\angle{PAQ}\equiv\angle{BAC}.$ Since $AR$ is also isogonal to the $A$-median with respect to $\angle{BAC}$, we obtain that $AO$ passes through the midpoint of $BC$, as desired.

Here's a sketch: Let $X = \odot(ABE) \cap \odot(ACD)$. $AX\perp PQ$, thus $AX$ and $AO$ are isogonal. $X$ is the centre of spiral similarity sending $CE$ to $DB$, thus, $X$ lies on the symmedian. Therefore, $AO$ is the median.

Let $A,X = (ABE)\cap (ACD)$. As $AX$ are radical axis of $(ABE)$ and $(ACD)$, $O_1O_2\perp PQ$. As $AX$ and $AO$ are isogonal with respect to $\angle BAC$, we need to show that $AX$ is $A-$ symmedian of $\triangle BAC$. $\sqrt{bc}$ inversion and reflection gives: $B$ and $C$ are points on sides $AD$ and $AE$ of triangle $ADE$ such that $BC||DE$. If $DC$ and $BE$ meet at $X$, then $AX$ is the $A-$ median of $\triangle ADE$ which is obvious by Ceva's.

Sketch of proof: It is sufficient to show that $\angle (O_1O_2, BA)=90^{\circ}-\angle CAM'$ where $M'$ is the midpoint of $\overline{BC}$. Claim. Line $\overline{O_1O_2}$ has fixed direction. (Proof) Animate $D$ on $\overline{AB}$ with constant velocity. Observe that $O_2 \mapsto D \mapsto E \mapsto O_1$ is a sequence of linear maps. Since $O_1 \equiv O_2$ when $D=B$, the line $\overline{O_1O_2}$ has fixed direction. So we need to solve the problem when $D=E=A$. This is easy trig.

Another solution Let $O_3$ be the circumcenter of $\triangle ABC$ we just proof $AO $,$CD$ and $BE$ are concurrent by Ceva. We simply know that $O_1O_3\perp AB$ and $O_2O_3\perp AC$ $\implies O_1O_3=\frac{\frac{1}{2}CE}{sinA}$ and $O_2O_3=\frac{\frac{1}{2}BD}{sinA}$ $\implies \frac{sin\angle BAM}{sin\angle CAM}=\frac{sin\angle O_3O_2O_1}{sin\angle O_3O_1O_2}=\frac{O_1O_3}{O_2O_3}=\frac{CE}{BD}=\frac{sinB}{sinC}$ and we get $AO ,CD$ and $BE$ are concurrent by Ceva.So we are done.

Let $X=(AEB)\cap (ADC)$ Since $AX\perp PQ$ we see that $AO$ and $AX$ are isogonal w.r.t $\triangle AQP \Longleftrightarrow$ $AO$ and $AX$ are isogonal w.r.t $\triangle ABC$. Hence it suffices to prove $AX$ is the $A-$symmedian of $\triangle ABC$. Performing $\sqrt{bc}$ inversion we see that $X^*=C^*D^*\cap B^*E^*$, which by ceva's theorem lies on the $A-$median in the inverted diagram. Inverting back we get the desired result.$\blacksquare$

Let $F=(ABE)\cap(ACD)$ and note $\angle BAF=90-\angle PQA=\angle MAC$ since $\overline{AF}\perp\overline{O_1O_2}.$ Take the $\sqrt{bc}$ inversion and we see $\overline{D^*E^*}\parallel\overline{B^*C^*}.$ Since $F^*=\overline{D^*C^*}\cap\overline{E^*C^*},$ we know $F$ lies on the $A$- median of the inverted triangle. Since $\overline{AM}$ is the isogonal conjugate of $\overline{AF},$ we conclude $\overline{AM}$ is the $A$-median of $\triangle ABC.$ $\square$

This is just Balkan MO 2009/2 lol.