In the quadrilateral $ABCD$ is $AD \parallel BC$ and the angles $\angle A$ and $\angle D$ are acute. The diagonals intersect in $P$. The circumscribed circles of $\vartriangle ABP$ and $\vartriangle CDP$ intersect the line $AD$ again at $S$ and $T$ respectively. Call $M$ the midpoint of $[ST]$. Prove that $\vartriangle BCM$ is isosceles.