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Found problems: 1

2021 Sharygin Geometry Olympiad, 10-11.6
Tags: geometry , trapezoid , angles , equa angles
The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle BSD$.