Let \(\overline{BD}\) and \(\overline{CE}\) be the altitudes of acute triangle \(ABC\). Let \(\omega_B\) and \(\omega_C\) be the circles with diameter \(\overline{BD}\) and \(\overline{CE}\), respectively. Suppose \(\overline{BD}\) intersect \(\omega_C\) at \(P\), \(\overline{CE}\) intersect \(\omega_B\) at \(R\), \(\omega_B\) and \(\omega_C\) intersects at \(X\) and \(Y\). If \(\overline{XY}\) intersect \(\overline{DE}\) at \(T\), prove that \(\overline{TP}=\overline{TR}\).