In isosceles $\triangle ABC$, $X$ is any point on $BC$. $P$ and $Q$ are the circumcentres of $\triangle ABX$ and $ACX$. Prove $APXQ$ is parallelogram.

Question

$X$ is any point on the base of an isosceles triangle $ABC$. $P$ and $Q$ are the circumcentres of the $\triangle ABX$ and $\triangle ACX$. Prove that $APXQ$ is a parallelogram.

My take on the question:
It is easy to prove that the figure $APXQ$ is a kite, from the fact that the diagonals are perpendicular to each other.
To prove that the figure is a parallelogram, we need to prove that the radii of both the circle are equal. If we are able to prove that points $P$ and $Q$ are equidistant from $AX$, then by congruence the radii can be proved equal.
I am not able to proceed any further. Can someone help me figure out the next step?

Answer 1

The base of the triangle is $BC$ right? And $X \in BC$? Since triangle $ABC$ is isosceles $$\angle \, ACX = \angle \, ACB = \angle \, ABC = \angle \, ABX = \beta$$

$$\angle \, AQX = 2 \, \angle \, ACX = 2 \beta$$ $$\angle \, APX = 2 \, \angle \, ABX = 2 \beta$$ Hence the two isosceles triangles $APX$ and $AQX$ are congruent which means $AP = AQ = XP = XQ$. The quad $APXQ$ is a rhombus.

Answer 2

The perpendicular bisector of AX passess through P and Q.

Draw the perpendicular bisectors of BX and XC, passing through P and Q respectively, and intersect them with a line through A that is parallel to BC, forming a rectangle. Convince yourself that turning this rectangle through 180° interchanges both A and X, and P and Q.

Here we have $BD=DX$ and $XE=EC$. Then

$$AF=DM=\tfrac12BX-MX = \tfrac12BM+\tfrac12MX-MX = \tfrac12MC-\tfrac12MX = \tfrac12XC = XE $$ So $AF=XE$ and rotating the rectangle by 180° interchanges A and X.