6. In a convex quadrilateral $ABCD$, angles $A$ and $C$ are both 100°. Points $X$ and $Y$ are chosen on sides $AB$ and $BC$ respectively such that $AX = CY$. It turns out that line $YD$ is parallel to the bisector of angle $ABC$. Find angle $AXY$. (A. Kuznetsov, S. Berlov)
Solution. Draw a line through point $Y$ parallel to $AB$. Let it intersect $AD$ at point $K$. Then $\angle DYC = \angle DYK$ and $\angle C = 100^{\circ} = \angle BAD = \angle YKD$, so triangles $DYC$ and $DYK$ are congruent by two angles and a side. Therefore, $YK = YC = AX$ and $AXYK$ is a parallelogram. Hence, $\angle AXY = \angle AKY = 80^{\circ}$.

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