The vertices of the pentagon $ABCDE$ are on a circle, and the points $H_1, H_2, H_3,H_4$ are the orthocenters of the triangles $ABC, ABE, ACD, ADE$ respectively . Prove that the quadrilateral determined by the four orthocenters is square if and only if $BE \parallel CD$ and the distance between them is $\frac{BE + CD}{2}$.