Let $ABC$ be an acute triangle and $T$ be a point interior to this triangle. that $\angle ATB = \angle BTC = \angle CTA$. Let $M,N$ and $P$ be the feet of the perpendiculars from $T$ to $BC$, $CA$ and $AB$ respectively. Prove that if the circle circumscribed around $\vartriangle MNP$ cuts again the sides $ BC$, $CA$ and $AB$ in $M_1$, $N_1$, $P_1$ respectively, then the $\vartriangle M_1N_1P_1$ It is equilateral.