Let $Z$ be the intersection of $\Gamma_1$ and $\Gamma_2$. Since $AD$ is common tangent line to $\Gamma_1$ and $\Gamma_2$,we know that $MZ$ cuts $AB$ at its midpoint. By Ceva's theorem we get that $SM$ cuts $AB$ at its midpoint thus $S,Z$ and $M$ are collinear so $S$ is on radical axes of two circles $\Gamma_1$ and $\Gamma_2$ so $SX \cdot SA= SY \cdot SB$ so $SO \perp XY$.

My solution: Let $ T $ be the midpoint of $ AD $ . Easy to see $ S, M, T $ are collinear . Since $ TA^2=TD^2 $ , so we get $ TM\equiv SM $ is the radical axis of $ \Gamma_1 $ and $ \Gamma_2 $ . Since $ SX \cdot SA=SY \cdot SD $ , so $ XY $ is anti-parallel to $ AD $ WRT $ \angle ASD $ , hence we get $ SO \perp XY $ . Q.E.D

My solution first of all it is well-known fact that if $c_{1}$ and $b_{1}$ are feets of perpendiculars from $C$ and $B$ to $AB$ and $CA$ then $AO$ is perpendicular to $B_{1}C_{1}$ so if we prove $XY$ is parallel to $B_{1}C{1}$ thanwe will get $ SO \perp XY $ it means we need to prove that $XY$ and $BC$ are anti-parallels. $\angle MAD=\angle MXA=\angle MCB$ it means $XBMC$ is cyclic then $\angle MDA=\angle MBD=\angle MBC$ it means $YCMB$ is cyclic so it means $XYCBM$ is cyclic so it means $XY$ is anti parallel to $BC$

My solution: WLOG, let $\angle CDA > \angle BAD $. Define: $N = \Gamma_1\cap \Gamma_2$ ($M $ and $N $ are distinct), $P = MN\cap AD $, $Q = XY\cap BC $, $E = PO\cap BC $, and $F = SO\cap XY $. Trivial angle chasing proves that $XYCMB $ is cyclic. This implies that $XYDA$ and $XYCB $ are also cyclic. From the Radical Axis Theorem, it follows that $S $, $N $, $M $ and $P $ are collinear. Also, power of a point implies that $P $ is the midpoint of $AD $. So, $OP\perp AD $ and $OE\perp BC $. It would now be enough to prove that $EFQO $ is cyclic. We have, $\angle FQE = \angle SXY - \angle SBC = \angle SDA - \angle SAD $. Also, $\angle EOS = 180 - \angle SOP = 180 - 2\angle SAD - \angle ASD = \angle SDA - \angle SAD $. Thus, $\angle EQF = \angle EOF $ and we are done! [NOTE: In my diagram $O$ lies in the interior od $ABCD $. If it lies in the exterior such that $\angle EOF $ is obtuse, then similar angle chasing will yield $\angle FOE + \angle FQE$ = $180$° and hence the result]. Edit: After nearly 9 months (while going through my posts), I now realize that I was dumb at that time (I didn't know directed angles). I should have directed the angles.

Let $\Gamma_1$ and $\Gamma_2$ intersect at $M$ and at another point $P$. Notice that since $AD$ tangent to $\Gamma_1$ at $A$ the following angle-chase holds $$\angle MXA = \angle MAD = \angle MCB \Rightarrow MCBX-cyclic$$similarly $MCBY$ is cyclic so we deduce that $XBCY$ is a cyclic quadrilateral Let line $t$ denote the tangent to the circumcircle of $\triangle ASD$ at $S$, obviously $SO \perp t$ $$\angle (t,SY) = \angle (t,SD)= \angle SAD = \angle SBC = \angle SYX \Rightarrow XY || t $$hence $$SO \perp XY$$

Obviously by angle chase we have that $MBYXC$ is a cyclic pentagon. (just notice that $\angle MAD = \angle MXA$ and $\angle MDA = \angle MYD$) Let $F$ be a point on $BC$ such that $SF \perp XY$, since $\angle CBX = \angle CYX$ this gives us that $\angle FSC = 90 - \angle CBS$, similarily we have that $\angle BSF = 90- \angle SCB$, this means that $SF$ and the $S$-altitude are isogonal lines w.r.t. $\angle BSC$, thus that means that $SF$ contains the circumcenter of $\triangle SBC$ (we call it $O'$). By homothety $\mathcal{H}$ centered at $S$ and which does the following: $B \overset{\mathcal{H}}{\rightarrow} A$ and $C \overset{\mathcal{H}}{\rightarrow} D$. This also means it takes $O' \overset{\mathcal{H}}{\rightarrow} O$. By colinearity of $S,O'$ and $O$ we have that $SO \perp XY$

Claim: $XYCMB$ is cyclic Proof: $$\angle{MYD}=\angle{MDA}=\angle{BDA}=\angle{DBC}=\angle{MBC},$$so $YCMB$ is cyclic, similarly, $XBMC$ is cyclic, so that gives the desired. $\Box$ To finish the problem, we just want to prove the tangent to $(ASD)$ at $S$ is parallel to $XY.$ From angle chasing we have $$\angle{CBS}=\angle{XYS}=\angle{DAS}=\angle{KSD},$$where $\angle{KSD}$ is the acute angle formed by $DS$ and the tangent to $(ASD)$ at $S.$ But that finishes the problem because $\angle{XYS}=\angle{KSD}.$ $\blacksquare$

Let N be midpoint of AD and G be second intersection of $\Gamma_1$ and $\Gamma_2$. we know AD is common tangent of $\Gamma_1$ and $\Gamma_2$ so G,M,N are collinear. also we know S,N,M are collinear so S lies on radical axis of $\Gamma_1$ and $\Gamma_2$ so ADYX is cyclic. Let L be line tangent to ASD at S clearly L || XY and we know SO ⊥ L so SO ⊥ XY. we're Done.