Let $K$ be the intersection of the perpendicular bisectors of $PX,QY$. Let $E,F$ be the midpoints of $PX,QY$. Let $EK,FK$ meet $BC$ at $M,N$. Then angle chasing gives $\angle PME = \angle KMD = \frac{\pi}{4}$, and symmetric angle chasing gives the same for $\angle KND$. Hence to finish, we just need to show $MB = NC$. Notice we can compute $MB = PB - PE \sqrt{2} = PB - \frac{PX}{\sqrt{2}} = PD - BD - \frac{PX}{\sqrt{2}}$. Hence we just need to show $\sqrt{2}(BD - CD) = QY -PX$. We can use power of a point to compute $PX = \frac{PB \cdot PC}{PA}$. Let $PD = x$, then we have $PX = \frac{(x - DB)(x +DC)}{x \sqrt{2}}$. Computing $QY$ symmetrically, we are done. In fact, we can show that $K$ is the intersection of the tangents from $B,C$. Sketch of Proof: Compute $AK^2 = EA^2 + EK^2 = EA^2 + AF^2$ and $AK'^2$ where $K'$ is the intersection of the tangents from $B,C$. Since both points lie on the bisector of $BC$ below $BC$, they must be the same.

The intersection of the perpendicular bisectors of $PX, QY$ is the midpoint of the segment connecting the intersection of the lines through $X, Y$ perpendicular to $AX, AY$ respectively, and the intersection of the lines through $P, Q$ perpendicular to $AX, AY$ respectively. The former point is the $A$-antipode (or the reflection of $A$ over circumcenter $O$), and the latter point is the reflection of $A$ over $D$. So, their midpoint clearly lies on the line through $O$ perpendicular to $BC$. $\square$

We will use coordinates. Let, $D:(0,0)$, $A:(0,a)$, $B:(b,0)$, $C:(c,0)$ By using problem condition, $P:(-a, 0)$, $Q:(a,0)$ Let the circumcircle of $ABC$ be $S=x^2+y^2+px+qy+r=0$ Using that $A$, $B$, $C$ lie on the circle, the equation becomes, $$S=x^2+y^2 - (b+c)x- \frac{a^2+bc}{a}y + bc=0$$To get coordinates of point $Y$, we solve, $x+y=a$ and $S$ $$Y= \left(\frac{a^2+ab+ac-bc}{2a} , \frac{a^2+bc-ab-ac}{2a} \right)$$$$T=\frac{Q+Y}{2}= \left( \frac{3a^2+ab+ac+bc}{4a}, \frac{a^2+bc-ab-ac}{4a} \right) $$As $AD=DQ$, $\angle DQA=45^{\circ}$, So, slope of $AQ=-1$. Hence, slope of perpendicular bisector of $QY=1$, and it passes through $T$. Call this line $L_1$. So, $$L_1: y- \left( \frac{a^2+bc-ab-ac}{4a} \right) = \left(x- \frac{3a^2+ab+ac-bc}{4a} \right)$$The line $L_1$ meets perpendicular bisector of $BC$ at $x=\frac{b+c}{2}$. Solving for $y=\frac{bc-a^2}{2a}$. So, point of intersection is $\left( \frac{b+c}{2}, \frac{bc-a^2}{2a} \right)$. This is symmetric in $b,c$, and we are done.

For a coordinate-based solution it's probably much more efficient to set $A=(0,1)$, so $P=(-1,0)$ and $Q=(1,0)$. Then we use power of a point to calculate $X$, since $PX\cdot PA=PB\cdot PC$ (where lengths are signed), which is equivalent to $PX\sqrt{2}=(-1-b)(-1-c)$ if I'm not mistaken.

One-Liner.

Evidently by dropping perpendiculars to $PX,QY$ we find $EX=EP, EY=EQ$, also observe $GHIJ$ is rectangle so drop perpendiculars from $G,E,I$ to $BC$, it suffices to show $BD=CD'$ which is plain obvious by $ABC \equiv A'CB$.

Oops. Denote by $M$ and $N$ the feet from $X$ and $Y$ and note that they lie on the perpendicular bisectors of $\overline{PX}$ and $\overline{QY}$ respectively. Thus it suffices to show $\overline{BM} = \overline{CN}$. This is a matter of length chasing. Note that from power of a point, \begin{align*} PB \cdot PC &= PX \cdot PA\\ \iff PB &= \frac{PX \cdot PA}{PC}\\ \iff MB &= \frac{PX \cdot PA}{PC} - PM\\ \iff MB &= \frac{PX \cdot PA}{PC} - \frac{PX}{\sqrt{2}}\\ \iff MB &= PX\left(\frac{PA\sqrt{2} - PC}{PC\sqrt{2}} \right)\\ \iff MB &= \frac{PX}{\sqrt{2}}\left(\frac{2AD - (PD + DC)}{PC} \right)\\ \iff MB &= \frac{PX}{\sqrt{2}} \left(\frac{QC}{PC} \right) \end{align*}Thus we have from similar calculations, \begin{align*} \frac{MB}{NC} &= \frac{PX}{QY} \cdot \frac{QC}{PC} \cdot \frac{QB}{PB}\\ &= \frac{PX}{QY} \cdot \frac{QY \cdot QA}{PX \cdot PA}\\ &= 1 \end{align*}so we are done. $\square$

Solve after a long time

Coordinates FTW! We employ cartesian coordinates with $B$ as the origin $(0,0)$ and the $x$-axis along the line $BC$. Let $A \equiv (b,c)$,$B \equiv (0,0)$ and $C \equiv (a,0)$.The coordinates of various points are as follows: $D \equiv (b,0)$ $P \equiv (b+c,0)$ $Q \equiv (b-c,0)$ This is because $DA=DP=DQ=c$. Now by applying power of point and finding the lengths it is not hard to see that: $X \equiv \left(\dfrac{(b+c)(a+c-b)}{2c},\dfrac{-(b+c)(a-b-c)}{2c}\right)$ and $Y \equiv \left(\dfrac{(b-c)(b+c-a)}{2c},\dfrac{(b-c)(b-a-c)}{2c}\right)$. Thus the equations of perpendicular bisectors of lines $BC,PX$ and $QY$ are as respectively: (Perpendicular Bisector of $BC$) $\boxed{x=\dfrac{a}{2}}$. (Perpendicular Bisector of $PX$) $\boxed{y=x+\left(\dfrac{b+c}{2c}\right)(b-c-a)}$. (Perpendicular Bisector of $QY$) $\boxed{y=-x+\left(\dfrac{b-c}{2c}\right)(b-a+c)}$. Clearly the perpendicular bisectors of $BC,PX$ and $QY$ are concurrent at $\left(\dfrac{a}{2},\dfrac{b^{2}-c^{2}-ab}{2c}\right)$. (QED) $\clubsuit$