Radical axes through a fixed point Given a convex quadrilateral ABCD with pairwise non­parallel sides. On the side AD an
arbitrary point P is chosen, different from A and D. The circumscribed circles of triangles
ABP and CDP intersect for the second time at point Q. Prove that the line P Q passes
through a fixed point, independent of the choice of the point P.
I tried to let I be the intersection of the radical axes of (ABP, CDP) and (ABP', CDP'). Then I wanted to show that I is on the radical axis of (ABE, CED) for any other point E on the line AD. But I have no idea how to relate the power of I to ABP, CDP, ABP', CDP' to its powers relative to ABE, CED.
Note: XYZ refers to the circumcircle of triangle XYZ.
 A: I do not see a proof using power of a point or radical axis theory, but here is a methodology that I found:
Hint 1:  See if you can find (by eye) some sort of possible relationship between $I$ and any of the other points.  This may help guide the solution.
Hint 2:

 Your diagram above is actually quite useful.  What property does quadrilateral $BQQ'C$ seem to have?  Can you prove it?

Solution:

 I claim that the locus of $Q$ as $P$ varies over $\overline{AD}$ is a fixed circle $\omega$ passing through $B$ and $C$.  To prove this, it suffices to show that the measure of $\angle BQC$ is fixed.  To wit, observe that $\angle CQP = 180 - \angle D$ and $\angle BQP = 180 - \angle A$ by the properties of cyclic quadrilaterals.  Hence $\angle BQC = 360 - \angle CQP - \angle BQP = \angle A + \angle D$ which is a fixed quantity.


 To finish, note that ray $\overline{PQ}$ intersects $\omega$ for the second time at a point $I$ which is fixed for all $P$.  This is because the location of $I$ on $\omega$ depends only on the measure of the arc $IC$ on $\omega$, which in turn depends only on the measure of $\angle IQC$.  But $\angle IQC = 180 - \angle CQP = \angle D$ which is fixed, hence so is $I$.

