Here is the solution for the location of points showed in the picture. For other cases, the location of the solution will be similar. Prove that for any selected point $D$ quadrilateral $XOYZ$ is inscribed. Then $\angle XZY = 180^{\circ} - 2\angle A$, so doesn't depend on $D$.

From the statement of the problem we have that circles $\triangle BXD$ and $\triangle DYC$ are tangent to lines $AB$ and $AC$. Denote these circles as $w_1$ and $w_2$, and their centers are $O_1$ and $O_2$. Let $K$ be the second point of intersection of these circles other than $D$ (see figure below).

Then
$$\angle BKC = \angle BKD + \angle DKC = \angle B + \angle C = 180^{\circ} - \angle A,$$
so $K$ is on the circumcircle $\triangle ABC$. Also we have
$$\angle XKY = \angle XKD + \angle DKY = \angle XBD + \angle DCY = 180^{\circ} - \angle XOY,$$
so quadrilateral $XOYK$ is inscribed. Because $AZ$ is the diameter of the big circle, we have that $AB \perp BZ$, so $O_1$ is on line $BZ$. Similarly $O_2$ is on the line $CZ$. We prove that $O_1$, $O_2$ are on the circle $OXKY$. Indeed,
$$\angle XO_1K = 2\angle XBK = 2(90^{\circ} - \frac{1}{2}\angle BOK) = 180^{\circ} - \angle XOK,$$
Similarly for point $O_2$. Now we prove that point $Z$ is on circle $OXKY$, which will finish the solution.

For this purpose we prove that $\angle O_1ZC + \angle O_1KO_2 = 180^{\circ}$. Indeed,
$$\angle O_1KO_2 = \angle O_1KD + \angle DKO_2 = 90^{\circ} - \angle DBK + 90^{\circ} - \angle DCK = 180^{\circ} - \angle A = 90^{\circ} - \angle BOC$$
$$\begin{align*} &= \angle BZC = 180^{\circ} - \angle O_1ZC \\ &\text{10th form} \\ &\text{10th form} \end{align*}$$