Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.
My progress: This problem is really intimidating to me !
I observed that XBY is collinear, which can be proved by angle chase. Just note that $\angle BDA = \angle CBA$ and $\angle ACB = \angle ABD $ . Then $\Delta ABD \sim \Delta ACB$ . By cyclic quads, we get XBY collinear .
Then I was able to show $PO_1=PO_2$ by noticing that $\angle PO_1O_2 = 180- \angle DAB$ and $\angle O_1O_2P = 180-\angle BAC$ .
Then I am stuck. I also observed that $O_1,P,O_2,Q$ is cyclic but was not able to prove.
I am also thinking of using spiral symmetry but I don't have any idea on how to use it ?
Please if possible send hints rather than solution. It helps me a lot . Thanks in advance.