Let the incircle of $\triangle ABD$ be tangent to $BC$ at $T$ and the incircle of $\triangle ADC$ be tangent to $BC$ at $S$. The radii of incircles of $\triangle ABD$ and $\triangle ADC$ are $r_1$ and $r_2$, respectively. Then
$$\begin{aligned} DS &= \frac{AD + DC - AC}{2} = \frac{AD + BC - BD - AC}{2} \\ &= \frac{AD - BD - AB + AB + BC - AC}{2} \\ &= \frac{AB + BC - AC}{2} - \frac{AB + BD - AD}{2} \\ &= BD - BT = TD. \end{aligned}$$
Since $X$ and $Z$ are centers of incircles, $XD \perp DZ$. Therefore,
$$XZ^2 = XD^2 + DZ^2 = r_1^2 + TD^2 + r_2^2 + TD^2.$$
and since
$$XZ^2 = (r_1 - r_2)^2 + TS^2 = (r_1 - r_2)^2 + 4TD^2$$
we have
$$\begin{aligned} r_1^2 + r_2^2 + 2TD^2 &= (r_1 - r_2)^2 + 4TD^2 \\ 2r_1r_2 &= 2TD^2. \end{aligned}$$
Thus,
$$\begin{aligned} XZ^2 &= r_1^2 + r_2^2 + 2TD^2 = (r_1 + r_2)^2 \\ XZ &= r_1 + r_2. \end{aligned}$$
Obviously, $XK \ge r_1$ and $ZK \ge r_2$. Since $XK + ZK = r_1 + r_2$, $XK = r_1$, $ZK = r_2$ and $XZ \perp AD$. Let $O$ be the center of circumscribed circle of $\triangle ABC$ (the midpoint of the segment $AC$). Then $OM \perp UV$. Since $AC$ is a radius $CY \perp AY$. Thus, $UV$ and $CY$ are parallel. Suppose $OM \cap CY = N$. Then $ON \perp CY$ and $YN = NC$. In the rectangle $KYNM$, $MK = NY$. Therefore, $CY = 2MK$. Done.