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 $\angle BXD - \angle BYD = \angle BSD$.
Note that $Y$ is unique given $X$. Now let $Y'$ be s.t. $S-X-Y'$ and $SX.SY'=SA.SC$. $AD\parallel BC$. So $\frac{SA}{SB}=\frac{SD}{SC}$ ie $SA.SC=SB.SD=SX.SY'$
Now $\frac{SX}{SC}=\frac{SA}{SY'}$ and $\angle XSC=\angle ASY'$. Hence $\triangle SXC\sim\triangle SAY'$. Similarly we obtain $\triangle SXB\sim\triangle SDY'$, $\triangle SXA\sim\triangle SDY'$ and $\triangle SXD\sim\triangle SAY'$.
So $\angle AXC-\angle AY'D=\angle XSA+\angle XAS+\angle XCS+\angle XSC-\angle AY'S-\angle DY'S=(\angle XSA+\angle XSD)+(\angle XAS-\angle DY'S)+(\angle XDS-\angle AY'S)=\angle ASC$. So $Y'=Y$.
By similar calculations we get $\angle BXD-\angle BY'D=\angle BSD$. But $Y'=Y$ as shown above. So $\angle BXD-\angle BYD=\angle BSD$ as desired.
Q.E.D.