Miquel theorem tells us,that if you select 3 points $X,Y$ and $Z$ on $CB,AC$ and $AB$, the cicumcercles of $\triangle XYC,\triangle XZB$ and of $\triangle YZA$ will met in one point.So we now that the third quadrilateral is also cyclic,because the first two meet in a point on the side of triangle and the second point is $D$,we conclude that the circumcercle of $\triangle XCY$ will pass through $D$. So we are done.

Something with simple tools: We suppose that the quadrilaterals $DXBZ,DXCY$ are cyclic and we will prove that $AZDY$ is also cyclic.Its enough to prove that $\angle{ADZ}=\angle{AYZ}$ but we have: $AZ\cdot AB=AD\cdot AX=AY\cdot AC$. So $BZYC$ is also cyclic and we get:$\angle{AYZ}=\angle{B}=\angle{ADZ}$ and hence the result.We work similarly for the other two cases. Best Regards, Christos

Unfortunately a circumscribable quadrilateral is not a cyclic quadrilateral, but a quadrilateral that can be circumscribed around a circle (i.e. contains an incircle) Suppose that the quadrilateral AZDY is circumscribable. Let the point of tangencies of the incircle of AZDY and sides AY, YD, DZ, ZA are P, Q, R, S, respectively. By straight computation we get $QY = \frac{BY + AY -c}{2}$ and $RZ=\frac{CZ+ AZ-b}{2}$ (where we are using $b=AC$ and $c=AB$) Therefore, we have $DQ=BY-QY -BD=\frac{BY-AY+c- 2BD}{2}$ and $DR = CZ-RZ-CD = \frac{CZ - AZ+b -2CD}{2}$. These two are equal, so $BY-CZ +AZ-AY = b-c+2(BD-CD)$. Using the fact that $AZ+ DY = AY+DZ$ (Pithot's theorem) we get that $c-b = BD-CD$. The converse of this is easily proven by assuming that the point D is such so that $c-b=BD-CD$, then introducing a point D' and Y' such that the line DD'Y' is tangent to the incircle of ACZ and that D' and Y' are on CZ and AC, respectively, and then showing that D is the same point as D'. Now, suppose that AZDY and CYDX are both circumscribable. Using the result above, we get that $DA- DC = AB-BC$, which proves that BXDZ is circumscribable.

suppose that quadrilaterals $DXBZ , DYAZ$ are cyclic.Thus , $\angle{YDX}$$=360-$$\angle{YDZ}$$-$$\angle{ZDX}$$=360-(180-$$\angle{A}$$)-(180-$$\angle{B}$$)$$=$$\angle{B}$$+$$\angle{A}$ Since $\angle{A}$$+$$\angle{B}$$+$$\angle{C}$$=180$ we get that $DXCY$ is cyclic as well

By Pitot's theorem on the 'concave' quadrangles $BCAD, ABCD$, we have that $DA-BC = DB-CA = DC-AB$, which gives that $ABDC$ has an incircle by the converse of the same theorem. As a side note, the point $D$ in the problem is the Outer Soddy point of $ABC$, also known as the Isoperimetric point and $X_{175}$ in ETC. It is the center of the outer Soddy circle, i.e. the circle that touches the circles centered at the vertices and orthogonal to the incircle internally.