KöMaL 2020 C1605

The diagonals of a convex quadrilateral $ABCD$ intersect at $M$. The area of triangle $ABM$ is greater than the area of triangle $CDM$. The midpoint of side $BC$ of the quadrilateral is $P$, and the midpoint of side $CD$ is $Q$, $AP+AQ=\sqrt2\,$. Prove that the area of quadrilateral $ABCD$ is less than 1.