2000 IMO Shortlist S19

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 $\measuredangle AYB = 2\cdot\measuredangle ADX$ .

设 $ABCD$ 为凸四边形。其边 $AB$ 和 $CD$ 的垂直平分线在 $Y$ 处相交。用 $X$ 表示四边形 $ABCD$ 内的一个点，使得 $\measuredangle ADX = \measuredangle BCX < 90^{\circ}$ 和 $\measuredangle DAX = \measuredangle CBX < 90^{\circ}$ 。表明 $\measuredangle AYB = 2\cdot\measuredangle ADX$ 。