Problems from Ukrainian Authors

Let $ABC$ be a triangle with orthocenter $I$, $D$ is the antipode of $I$ in $(BIC)$ (we'll call this circle as $\omega$). So, $ABCD$ is a parallelogram. A circle, which passes through the points $A$ and $I$, intersects $\omega$ at $X$ and lines $BI$ and $CI$ at points $Y$ and $Z$ respectively ($X,Y,Z\ne I$). Lines $AY$ and $AZ$ intersect $(ABC)$ for the second time at points $U$ and $V$ respectively. Prove that the points $D$, $U$, $V$ and $X$ are concyclic. Let line $UV$ meets $BC$ at point $L$. Then it will be sufficient to prove that $DX$ passes through the point $L$ (because then $LU\cdot LV=LB\cdot LC=LX\cdot LD$ and the problem will be solved). Let $K$ be the second intersection point of $(ABC)$ and $AX$. Observe that (fig. 22)

$$\angle (VA,AU)=\angle (ZA,AY)=\angle (ZI,IY)=\angle (CI,BI)=\angle (BA,AC),$$ So, arcs $UV$ and $BC$ are equal and then $UV=BC$. Triangles $UVK$ and $BCX$ are similar, because $$\angle UVK=\angle UAK=\angle YAX=\angle (XI,IY)=\angle (XI,IB)=\angle (XC,CB)=\angle BCX$$ and, similarly, $\angle VUK=\angle CBX$.

So, points $D$, $L$ and $X$ are collinear.