Školjka

%V0 Let $ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $AB$ and $CD$ meet at $Y$. Denote by $X$ a point inside the quadrilateral $ABCD$ such that $\measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $\measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $\measuredangle AYB = 2\cdot\measuredangle ADX$.