The inscribed circle $\omega$ of an equilateral $\Delta ABC$ is tangent to its sides $AB$,$BC$ and $CA$ in points $D$,$E$, and $F$, respectively. Point $H$ is the foot of the altitude from $D$ to $EF$. Let $AH\cap BC=X,BH\cap CA=Y$. It is known that $XY\cap AB=T$. Let $D$ be the center of the circumscribed circle of $\Delta BYX$. Prove that $OH\perp CT$.