The quadrilateral $ABCD$ is inscribed in a circle. The lines $AB$ and $CD$ meet each other in the point $E$, while the diagonals $AC$ and $BD$ in the point $F$. The circumcircles of the triangles $AFD$ and $BFC$ have a second common point, which is denoted by $H$. Prove that $\angle EHF=90^\circ$.