The key observation of this problem is to consider the intersection of the lines $XY$ and $ST$. Suppose $DS$, $DT$ intersect $\Omega$ again at $P$, $Q$, and $BQ\cap CP=R$. By applying Pascal's theorem on $PDQBAC$ and $PDQBEC$, we have $XY\cap ST=R$. Claim. $C, R, Y, T$ are concyclic. Proof: With the condition that $AD$ is tangent to $\odot(DXY)$, we have the following angle chasing \[\measuredangle TCR=\measuredangle ACP=\measuredangle ADP=\measuredangle ADX=\measuredangle DYX=\measuredangle TYR.\] $AE$ is tangent to $\odot(EXY)$ $\iff \measuredangle XYE=\measuredangle XEA$. \[\measuredangle XYE=\measuredangle RYC = \measuredangle RTC = \measuredangle BCA = \measuredangle BEA = \measuredangle XEA, \]so we are done.