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 $KA^2 = KM \cdot KL = KD^2$ (or $KA^2 = KM^2 = KD^2$) $\Rightarrow$ $KA = KD$.

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,
$$\begin{aligned} \angle SYX + \angle YSO &= \angle SAD + \angle YSO = 0.5\angle SOD + \angle YSO = \\ &= 0.5(180^\circ - \angle OSD - \angle ODS) + \angle YSO = [\angle OSD = \angle ODS] = \\ &= 0.5(180^\circ - 2\angle ODS) + \angle YSO = 90^\circ + (\angle YSO - \angle ODS) = 90^\circ, \end{aligned}$$
which implies $SO \perp XY$.