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 ($X,Y,Z \neq I$). 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.