60th Belarusian Mathematical Olympiad

Draw the line $LM$, where $L$ is the second (different from $M$) point of intersection of ${\Gamma }_1$ and ${\Gamma }_2$. (If ${\Gamma }_1$ touches ${\Gamma }_2$ at $M$, then $LM$ is the common tangent of these circles at $M$.) Let $K$ be the point of intersection of the lines $LM$ and $AD$. By the tangent-secant theorem, we have $K{A}^2=KM\cdot KL=K{D}^2$ (or $K{A}^2=K{M}^2=K{D}^2$) $\Rightarrow$ $KA=KD$.



It is well-known fact that the midpoint of the base, of the trapezium, $K$, the point of intersection of the trapezium diagonals, $M$, and the point of intersection of the extensions of its lateral sides, $S$, lie on the same line. Then $SX\cdot SA=SL\cdot SM=SY\cdot SD$. Hence $SX:SY=SD:SA$, which gives the similarity of the triangles $SXY$ and $SDA$. Thus $\angle SYX=\angle SAD$. The angles $SOD$ and $SAD$ are respectively a central and an inscribed angle of the same circle, and both of them are subtended by the same chord $SD$. Hence $\angle SAD=0.5\angle SOD$. Therefore, $$\begin{array}{rl}\angle SYX+\angle YSO& =\angle SAD+\angle YSO=0.5\angle SOD+\angle YSO=\\ & =0.5(180^{\circ }-\angle OSD-\angle ODS)+\angle YSO=[\angle OSD=\angle ODS]=\\ & =0.5(180^{\circ }-2\angle ODS)+\angle YSO=90^{\circ }+(\angle YSO-\angle ODS)=90^{\circ },\end{array}$$ which implies $SO\perp XY$.