See that $\angle DCX = \angle ABX=\angle XBY = \angle XCY$ Hence, $Y$ lies on $CD$. Now $\angle XYD = \angle XBC = \angle XAD = \angle XYC$. Since $Y \in CD$, we conclude that $\angle XAD=90^\circ$. Hence $\angle CAD=90^\circ$. $\blacksquare$.

Claim.1 : $Y \in CD$(like Synthetic_Potato): We have:$$\angle{YCX}=\angle{YBX} =\angle{XBA} =\angle{DCX} $$$$\Longrightarrow Y \in CD$$ Claim.2 : $\angle{DYX} =90$: Let $K=AB \cap DC$. Notice that $Y$ is the miquel point of the cyclic quadrilateral $ACBD$. $\Longrightarrow \angle{KYX} =90 \Longrightarrow \angle{DYX} =90 \Longrightarrow \angle{CAD} =90$. As desired... $\blacksquare$

$ \angle YAB=2m$ for some m,Notice, $\angle YXB= 90+ .5( \angle YAB)=90+m. $In triangle YAB . $\angle YXD=180- \angle YXB=90-m$ .As ABCD is cyclic,$\angle CAB=\angle CDB=m.$ In triangle YDX$ ,\angle CYX=\angle CDX +\angle DXY= m+ 90-m=90.$ But,$\angle CYX =\angle CAD(cyclic quad.)$ hence proved

Join $XY$ and let $\angle AYB=2a$, $AXYD$ is cyclic $\implies \angle AYX=\angle ADX=a$.Also $\angle AXB=90+a \implies \angle AXD=90-a$.So in $\Delta AXD,\angle XAD=180-90+a+a=90=\angle CAD$