Turkish Mathematical Olympiad

Let the incircle of $△ABD$ be tangent to $BC$ at $T$ and the incircle of $△ADC$ be tangent to $BC$ at $S$. The radii of incircles of $△ABD$ and $△ADC$ are $r_1$ and $r_2$, respectively. Then $$\begin{array}{rl}DS& =\frac{AD+DC-AC}{2}=\frac{AD+BC-BD-AC}{2}\\ & =\frac{AD-BD-AB+AB+BC-AC}{2}\\ & =\frac{AB+BC-AC}{2}-\frac{AB+BD-AD}{2}\\ & =BD-BT=TD.\end{array}$$ Since $X$ and $Z$ are centers of incircles, $XD\perp DZ$. Therefore, $$X{Z}^2=X{D}^2+D{Z}^2=r_1^2+T{D}^2+r_2^2+T{D}^2.$$ and since $$X{Z}^2=(r_1-r_2{)}^2+T{S}^2=(r_1-r_2{)}^2+4T{D}^2$$ we have $$\begin{array}{rl}{r}_1^2+r_2^2+2T{D}^2& =(r_1-r_2{)}^2+4T{D}^2\\ 2r_1{r}_2& =2T{D}^2.\end{array}$$ Thus, $$\begin{array}{rl}X{Z}^2& =r_1^2+r_2^2+2T{D}^2=(r_1+r_2{)}^2\\ XZ& =r_1+r_2.\end{array}$$ 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 $△ABC$ (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.