Denote by $C_1$ and $C_2$ the circumcircles of the triangles $ABC$ and $DEF$, respectively.
Since $\angle PAM = \angle PDM = 60^\circ$, the quadrilateral $APMD$ is cyclic, so $\angle APD = \angle AMD = x^\circ$.

Looking at the quadrilateral $APIM$ we have: $\angle PIM = 360^\circ - \angle MAP - \angle AMI - \angle API$, so $\angle PIM = 360^\circ - 60^\circ - 2 \cdot \frac{180^\circ - x^\circ}{2} - x^\circ = 120^\circ$.
Since $\angle PIM = 180^\circ - \angle PAM$, we infer that $APIM$ is a cyclic quadrilateral, so $ADIP$ and $ADMI$ are cyclic too.

It follows that $\angle DAI = \angle DPI = \frac{180^\circ - x^\circ}{2}$ and $\angle ADI = \angle AMI = \frac{180^\circ - x^\circ}{2}$.
Subsequently, $\angle ADI = \angle DAI$, so $AI = DI$.
Let $A'$ be the second intersection point of $AI$ and $C_1$ and $D'$ be the second intersection point of $DI$ and $C_2$.
Since $ADIP$ is cyclic, it follows that $\angle IAP = \angle IDP$, so the arc $A'C$ from the circle $C_1$ and the arc $D'F$ from the circle $C_2$ have the same measure. Since the minor arcs $AC$ and $DF$ from the two circles are congruent, each having $120^\circ$, it follows that the arcs $ACA'$ (in the circle $C_1$) and $DFD'$ (in the circle $C_2$) have the same measure, so $AA' = DD'$.

From here, we obtain $A'I = AA' - AI = DD' - DI = D'I$, so $IA \cdot IA' = ID \cdot ID'$, which means that the power of the point $I$ with respect to the circle $C_1$ is equal to the power of $I$ with respect to $C_2$. Hence, $I$ belongs to the radical axis $UV$ of these two circles.
Analogously, we prove that $J \in UV$, so the lines $UV$ and $IJ$ coincide. Hence, $IJ$ is the radical axis of the circles $C_1(O_1)$ and $C_2(O_2)$, therefore is the perpendicular bisector of $O_1O_2$.