Point $T$ lies on the bisector of $\angle B$ of acuteangled $\triangle ABC$. Circle $S$ with diameter $BT$ intersects $AB$ and $BC$ at points $P$ and $Q$. Circle, that goes through point $A$ and tangent to $S$ at $P$ intersects line $AC$ at $X$. Circle, that goes through point $C$ and tangent to $S$ at $Q$ intersects line $AC$ at $Y$. Prove, that $TX=TY$