Let $AO$ intersect $\Omega$ at $F$. Clearly $AB\parallel CF$ so we must have $(A,B;E,F)=(CA,CB;CE,CF)=(A,B;N,P_{\infty})=-1$ where $P_{\infty}$ is the point at infinity along line $AB$. It follows that $XF$ is tangent to $\Omega$ at $F$. Similarly, $YF$ is tangent to $\Omega$ at $F$, so we are done.

Let the other tangency point from $Y$ be $E'$ and the midpoint of $AB$ be $M$. Let the intersection of $E'O$ with $BA$ be $A'$. Note that $\angle OMC=\angle MCA=\angle MBE$ and $\angle OEM=\angle OCE=\angle BEY$ so $\triangle OME\sim \triangle YBE \sim \triangle YEA$. Hence $\sin\angle OA'B=\frac{YE'}{YA}=\frac{YE}{YA}=\frac{OM}{OE}=\frac{CA}{CB}=\sin\angle OAB$. It is easy to see we must have $\angle OA'B=\angle OAB$, hence $E'$ is the point diametrically opposite $A$. This means the second tangency point from $X$ is also diametrically opposite $A$, hence $XY$ is tangent to $\omega$.

Quite easy. Let $Q$ be the midpoint of $BC$ and $AQ$ meets $(ABC)$ at $P$, so we will prove that $AEBP$ is harmonic, which is obviously sufficient. We will prove that $EP$ is symmedian for $\triangle ABE$. Indeed, this is just angle chase: $\angle BEP= \angle BAP =\angle ABC =\angle AEC$, so we are done.

Projective cookie. Let $T$ be point such that $BACT$ is parallelogram. Since $\angle CAB = \angle CTB=90^{\circ}$ point $T$ obviously lies on $\odot(ABC)$. Moreover, $CT \parallel AB$ and $BT \parallel AC$. Note that: $$ -1=(A,C;M, \infty) \overset{B}{=} (A,C;D,T) \quad \text{and} \quad -1 =(A,B;N,\infty) \overset{C}{=} (A,B;E,T)$$This means that quadrilaterals $AEBT$ and $ADCT$ are harmonic; therefore tangents at points $E,T$ to $\odot(ABC)$ intersects on the line $AB$ and tangents at the points $D,T$ to $\odot(ABC)$ intersects on the line $AC$. We conclude that points $X,T,Y$ are collinear and line $XY$ is tangent to $\odot(ABC)$ at the point $T$.

Attachments:

Too easy to guess the point (which is the antipode A' of A) and to bash(one can easily prove that $AC \cdot AX =AA'^2$ by calculating $AX$) so I think this is not a good problem (nice result though).

Accurate diagram = 99% of the solution. (And I really appreciate that simple similarities suffice and in particular one does not need to operate with harmonic craziness.) Let $A'$ be the point diametrically opposite point to $A$ in $\Omega$ - if we show that $YA'$ is tangent to $\Omega$, i.e. $\angle OA'Y = 90^{\circ}$ (where $O$ is the midpoint of $BC$ and hence the center of $\Omega$), then it will analogously follow that $XA'$ is tangent to $\Omega$ and we will be done. Equivalently, it suffices to have $\angle AA'Y = 90^{\circ}$, i.e. (since $ABA'C$ is a rectangle) $AA'^2 = AB \cdot AY$. Let $M$ be the midpoint of $AB$ and $R$ be the radius of $\Omega$ - we want $AY = \frac{4R^2}{AB}$. We have $\angle MEO = 90^{\circ} - \angle BEO = \angle YEB = \angle YAE$ and $\angle EMO = 90^{\circ} + \angle EMB = 180^{\circ} - \angle EBM = \angle EBY = \angle AEY$, thus $\triangle EMO \sim \triangle AEY$ implying $\frac{EO}{AY} = \frac{EM}{AE}$. Hence $AY = \frac{EO \cdot AE}{EM} = R \cdot \frac{AE}{EM} = R \cdot \frac{BC}{BM} = \frac{2R^2}{AB/2} = \frac{4R^2}{AB}$, as required!

Let $G$ be the point such that $ABGC$ is a rectangle and let $M,N$ the midpoints of $AC,AB$ respecitivily, projecting gives $$-1=(A, B; N, \infty_{AB}) \overset{C}{=} (A, B; E, G) \implies GX \; \text{tangent to} \; \Omega$$$$-1=(A, C; M, \infty_{AC}) \overset{B}{=} (A, C; D, G) \implies GY \; \text{tangent to} \; \Omega$$Both mean that $XY$ is tangent to $\Omega$ thus we are done

Fortunately, we indeed do not need any projective geometry and just a minimal amount of length bashing: Let $M_A, MB_B, M_C$ be the three midpoints of $ABC$. By Thales, $M_AM_CEY$ is cyclic. Let $A'$ be the antipode of $A$ on the circle. We prove that $M_AM_CEA'$ is cyclic. Indeed, we have $$SM_C \cdot SE= \frac{1}{2} SC \cdot SE= \frac{1}{2} SA \cdot SA'=SM_A \cdot SA'$$where $S$ is the centroid, proving our claim. Thus, we have shown that $A'Y$ is tangent to $\Omega$ at $A'$ and since the same holds for $A'X$ by symmetry, we are done.

It can be solved by angle chasing too (no new points), so easy. Let $M$, $N$, $O$ be the midpoints of $AB$, $AC$, $BC$ respectively and $L$ be the foot of perpendicular from $O$ to $BC$. Since $\angle ONX=\angle ODX=\angle OLX=90$ we have $(O N D X L)$ is cyclic, similarly $(O M E Y L)$ is cyclic. Rest is angle chasing. Obtain that $\angle DLE=\angle OLD+\angle OLE=\angle ONB+\angle OMC=90-\angle MCB-\angle NBC$, and it is obvious that $\angle DOE=180-2\angle MCB-2\angle NBC$, which means that $2\angle DLE=\angle DOE$, adding to this $OE=OD$, gives that $O$ is circumcenter of $DLE$, then $L$ lies on $(A B C)$, since $OL$ is perpendicular $XY$, we have $XY$ tangent to $(A B C)$, so we are done:).