First, we claim that triangles \( \triangle ABC \) and \( \triangle AED \) are similar. To prove this, observe that \[ \frac{AB}{BC} = \left(\frac{AB}{BX}\right) \div \left(\frac{BC}{BX}\right) = \left(\frac{EY}{ED}\right) \div \left(\frac{EY}{AE}\right) = \frac{AE}{ED}. \]Since \( \angle ABC = \angle AED \), the similarity follows. Thus, we deduce that \( \angle CAX = \angle DAY \). This is sufficient to show that the circumcircles of triangles \( \triangle ACD \) and \( \triangle AXY \) are tangent, one way is inversion at \( A \). (If someone knows a name for this tangency lemma please let me know).

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Denote by $[\triangle ABC]$ as the area of $\triangle ABC$. Note that by $\triangle EDY \sim \triangle YEA$ we have $\frac{ED}{DY}=\frac{EY}{YA}$ so $ED \cdot EA = EY \cdot DY$. $$[\triangle EDY]=\frac{\sin(\angle EYD) \cdot EY \cdot DY}{2} = \frac{ \sin(\angle YED + \angle EDY) \cdot ED \cdot EA}{2} = \frac{\sin(\angle YED + \angle AEY) \cdot ED \cdot EA}{2} = [\triangle EDA] $$Similarly, $[\triangle CBA]=[\triangle CBX]$. $$(\frac{ED}{BC})^2= \frac{[\triangle EDY]}{[\triangle CBX]}=\frac{[\triangle EDA]}{[\triangle CBA]}=\frac{ED\cdot EA \cdot \frac{sin(\angle AED)}{2}}{CB\cdot BA \cdot \frac{sin(\angle ABC)}{2}} = \frac{ED\cdot EA}{CB\cdot BA}$$We get $\frac{EA}{ED}=\frac{BA}{BC}$. So, $\triangle AED \sim \triangle ABC$. Hence, $\angle CAX = \angle DAY$ which shows that circumcircles of $\triangle ACD$ and $\triangle AXY$ are tangent.