Extend $AX, AY$ to meet $(ABC)$ again at $P,Q$. Since $\angle XPB = \angle APB = \angle ACB = \angle XDB$, we have $P \in (BDE)$. Similarly, $Q \in (CDE)$. Let $AH$ extended meet $(ABC)$ at $Z$. Observe that $HE.HD = (2HA)(\frac{HZ}{2}) = HA.HZ$, so $H$ must be the radical center of $(ABC), (BDE), (CDE)$. So $B,H,P$ and $C,H,Q$ are collinear. Note that this means $P,Q$ are reflections of $H$ across sides so $AP = AQ$. To finish, see that $\angle XAQ = 2 \angle APQ = \angle XYA + \angle ACQ$, so the circles are indeed tangent, as desired. $\blacksquare$

Let $BH$ and $CH$ meet circle of $ABC$ at $S,K$. Claim $: BDSE$ and $CDKE$ are cyclic. Proof $:$ Note that $BH.HS = AH.2HD = 2AH.HD = EH.HD \implies BDSE$ is cyclic. we prove the other one with same approach. Claim $: X,A,S$ and $Y,A,K$ are collinear. Proof $:$ Note that $\angle XSB = \angle XDB = \angle ACB = \angle ASB \implies X,A,S$ are collinear. we prove the other one with same approach. Note that $\angle YAS = \angle YXA + \angle XYA$ and we need to prove $\angle YAS = \angle YXA + \angle AKS$ so we need to prove $\angle AKS = \angle XYA$. Claim $: XYSK$ is cyclic. Proof $:$ Note that $KA.AY = EA.AD = SA.AK$. Now Note that we need to prove $AK = AS$ which is true since $AK = AH = AS$.

Let $D'$ be the reflection of point $D$ respect to $A$. So $ED'=HD$ and since $XEDB$ is cyclic , $\angle XBD=\angle XED'$. By $\angle XEB=\angle XDB=\angle C$ one can see that : $\frac{XE}{XB}=\frac{HD}{BD}=\frac{ED'}{BD}$ So the triangles $XED'$ and $XBD$ are similar and $\angle D'XD=\angle EXB=90$. So $AD'=AD=AX$ and similarly $AY=AD$ , So $A$ is the circumcenter of triangle $DXY$ And easily by angle chasing one can see that $\angle XAB=\angle ACB+\angle AYX$ , which implies circumcircle of tringles $ABC$ and $AXY$ are tangent at $A$ , as desired.

Let the reflections of $H$ over $BC, CA, AB$ be $H_A, H_B, H_C$ respectively. The Orthocenter Reflection Lemma implies that all $3$ points lie on $(ABC).$ Now, observe $$HD \cdot HE = \frac{HH_A}{2} \cdot (2 \cdot HA) = HA \cdot HH_A$$$$= | Pow_{(ABC)}(H) | = HB \cdot HH_B$$so $H_B$ lies on $(BDEX)$. Because $AC \parallel DX$, we know $A \in H_BX$ follows from Reim's. An analogous process yields $A \in H_CY$. Because $A$ lies on $DE$, which is the Radical Axis of $(BDE)$ and $(CDE)$, we have $$AH_B \cdot AX = | Pow_{(BDE)}(A) | = | Pow_{(CDE)}(A) | = AH_C \cdot AY.$$But properties of reflections give $AH_B = AH = AH_C$, so $AX = AY$ also holds. Now, the desired result follows readily. $\blacksquare$

Solved with Shreyasharma who had the smart idea of invoking homothety. Pretty easy for an Iran Geo. Let $H_B$ and $H_C$ be the reflections of the orthocenter across the sides $AC$ and $AB$ respectively. It is well known that $H_B$ and $H_C$ lie on $(ABC)$. Let $\omega_B$ and $\omega_C$ be the circles $(EDB)$ and $(EDC)$ respectively. We start off by making the following key observations. Claim : Points $H_B$ and $H_C$ lie on $\omega_B$ and $\omega_C$ respectively. Proof : We simply note that \[AH_B=AH=AE\]which implies that $A$ is the center of the circle $(EH_BH)$. Thus, it follows that \[\measuredangle BH_BE = \measuredangle HH_BE = 90^\circ = \measuredangle ADE\]which proves that $H_B$ lies on $\omega_B$ as claimed. Similarly, it follows that $H_C$ lies on $\omega_C$ as well. Claim : Points $X$, $A$ and $H_B$ are collinear. Similarly, points $Y$, $A$ and $H_C$ are also collinear. Proof : Let $X_1 = \overline{H_BA} \cap \omega_B$. Then, since we have already showed that $AE=AH_B$, by similarly it follows that $AX_1=AD$ as well. Thus, \[2\measuredangle AX_1D = \measuredangle H_AAD = 2\measuredangle ACB\]Thus, $\measuredangle AX_1D = 90 + \measuredangle ACB = \measuredangle CAD$. But this implies that $X_1D \parallel AC$. Thus, $X_1=X$ which proves the claim. Similarly, we have $Y-A-H_C$ as well. Now, note that since $AX=AY$ and $AH_B=AH_C$, using the previous collinearity it follows that triangles $\triangle AXY$ and $\triangle AH_BH_C$ are homothetic. Thus, it is clear that $(AXY)$ and $(AH_BH_C)$ are tangent, from which it follows that the circumcircle of triangle $AXY$ is tangent to that of $ABC$. Remark : We also noticed that once you reflect the points $X$ and $Y$ across $A$ you came across all sorts of claim. For example, $X'$ and $Y'$ lies on $\omega_B$ and $\omega_C$ and are collinear with $AX$ and $AY$. Further, it was also interesting to note that if $P$ and $Q$ were the feet of the altitudes from $B$ and $C$, then $X'-P-Q-Y'$.

Took me long enough, also my solution is similar to others but I am going to post it for storage. Thanks JanHajfor suggesting me to try this problem Let $\odot (ABC)=\Omega$, $\odot(EDBX)=\omega_1$ , $\odot(EDCY)=\omega_2$. Let $CH \cap \Omega = \{P\}$ and $BH \cap \Omega=\{Q\}$ Claim: Points $P , Q \in \Omega$ Proof: Let $H'$ be the reflection of $H$ over the segment $BC$. It's well known that $H' \in \Omega$. Now by Power of the Point Theorem (POP) we get: $HP \cdot HC = Pow(H, \omega_1)=HE \cdot HD=\frac{HA}{2} \cdot 2 HH'=HA \cdot HH' \implies HP \cdot HC = HH' \cdot HP$ So by the converse of POP we get that points $A,P,H '$ and $C$ are concyclic $\iff P \in \odot (ACH') \implies P \in \Omega$ Similarly we get that $Q \in \Omega$ $\square$. Claim: Points $\overline{P-A-Y}$ and $\overline{Q-A-X}$ are collinear. Proof: Let $PA \cap \omega_2=\{Y'\}$ and $QA \cap \omega_1=\{X'\}$ $\angle CPY' \equiv CPA \stackrel{\Omega}{=} \angle CBA \stackrel{AB \parallel DY}{=} \angle CDY \stackrel{\omega_2}{=} \angle CPY \implies \angle CPY' = \angle CPY \implies Y'=Y$ Hence points $\overline{P-A-Y}$ are collinear. Similarly points $\overline{Q-A-X}$ are collinear. $\square$ Claim: Points $X,P,Q$ and $Y$ are concyclic. Proof: Again by POP we get: $AP \cdot AY=Pow(A,\omega_2)=AE \cdot AD=Pow(A,\omega_1)=AQ \cdot AX \implies AP \cdot AY=AQ \cdot AX \therefore$ Points $X,P,Q$ and $Y$ are concyclic. $\square$ , let (XPQY)=$\Gamma$. Claim: $\odot (AXY)$ is tangent to $\odot ABC)$ Proof: Note that $\angle APQ= \angle ABQ=90-\angle A=\angle ACP=\angle AQP \implies APQ=\angle AQP \implies AP=AQ$. Since $AP=AQ$ our claim is equivalent to proving that the triangle $\triangle XAY$ is isosceles, but this is true since: $\angle AXY \equiv QXY \stackrel{\Gamma}{=} \angle QPY \equiv QPA \stackrel{AP=AQ}{=} \angle PQA \equiv PQX \stackrel{\Gamma}{=} \angle PYX \equiv AYX \implies \angle AXY=\angle AYX \implies AX=AY$ $\blacksquare$

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