$DPXR, DQYS$ are rectangles with diameters equal to $PR = QS = BD = DC$, so $DX = DY \implies \triangle{DXY}$ is isosceles, and $Z$ is the midpoint of $XY$. $\square$

Consider that $\angle XZD=90^{\circ}=\angle YZD$ Hence $X-Z-Y$ Consider that $XD=PR=BD=DC=SQ=DY$ Hence $XZ=YZ$

First, we know that $\angle DPA= \angle DQA= 90\implies \angle XZD=\angle YZD= 90$, because $(DPXZR)= \Gamma_1$ and $(DQYZS)= \Gamma_2$ are cyclic. Notice that with this, we have $X,Y,Z$ collinear and we want to prove $XD=DY$ (if this happens, $XYD$ will be isosceles $\implies XZ=ZY$). Also, as $D$ and $F$ are midpoints $\implies DF//AC\implies \angle SYA= \angle DSY= 180-\angle YQD= 180-90=90$. So, we have $\angle SYA= 90\implies \angle SYQ= 90$. (i) Similarly, $\angle RXP= 90$. (ii) By (i) and (ii), we now know that $PR$ and $SQ$ are the diameters of $\Gamma_1$ and $\Gamma_2$, respectively. And, as $\angle DQY= 90\implies YD$ diameter of $\Gamma_2$. Besides that, we know $\angle DPX= 90\implies XD$ diameter of $\Gamma_1$. So, $PR= XD$ and $SQ= YD$. Now, we just have to prove that $PR= SQ$ and the problem is finished! But, $EF//BC$ and, by definition, $PR//BC$ and $SQ//BC\implies$ $EF//PR (*)$ $EF//SQ(**)$ In (*), as $EF//PR$ and $ED//AB\implies ER//FP\implies FERP$ is parallelogram. Which means $EF= PR$. Similarly, in (**), as $EF//SQ$ and $DF//AC\implies FS//EQ\implies FEQS$ is parallelogram, so $EF= SQ$. Then, finally, we have $EF=PR=SQ$, as we wanted.

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??? $DPXR$ and $DQYS$ are rectangles, so $X$ and $Y$ are the antipodes of $D$ in their respective circles. Furthermore $BDRP$ and $CDSQ$ are parallelograms, hence $PR=QS$ so their circumcircles are congruent. The rest is obvious by symmetry.

$DPXR$ is a rectangle $\implies DX\parallel$ reflection of $BC$ over $DP\implies \triangle DBX$ isosceles $\implies X$ is the foot of the altitude from $C$, similarly $Y$ is the foot if the altitude from $B$ so $DX = DY$ and $Z = (DX)\cap (DY) = $ midpoint of $XY$.

Didn't not expect such a direct proof. $SQPR$ and $QSPR$ are parallelograms, $ YQDS$ and $XRPD$ are rectangles $\implies DX=PR=SQ=DY$ so $\triangle DXY$ is isosceles. $X,P,Z,R,D$ are concyclic on a circle with diameter $XD \implies \angle XZD=90^{\circ}$ therefore $DZ$ is perpendicular bisector of $XY$ so $XZ=ZY$

Because $AD$ is a diameter, we get $\angle DPA=\angle DPX=90^{\circ}\Rightarrow\angle DZX=90^{\circ}$, analogously $\angle DZY=90^{\circ}\Rightarrow Z\in XY.$ It suffices to show that $XD=YD$, but $\angle BXD=\angle PRD=\angle FED=\angle ABD\Rightarrow XD=BD=\frac{BC}{2}=YD$, analogously.