PL-OM 2007 Problem 5

In tetrahedron $ABCD$ following equalities hold: $\angle BAC+\angle BDC=\angle ABD+\angle ACD$ $\angle BAD+\angle BCD=\angle ABC+\angle ADC$ Prove that center of sphere circumscribed about ABCD lies on a line through midpoints of $AB$ and $CD$ .