In unequal triangle $ABC$ bisector $AK$ was drawn. Diameter $XY$ of its circumcircle is perpendicular to $AK$ (order of points on circumcircle is $B-X-A-Y-C$). A circle, passing on points $X$ and $Y$, intersect segments $BK$ and $CK$ in points $T$ and $Z$ respectively. Prove that if $KZ=KT$, then $XT \perp YZ$.