We show that the line through the centres of the circles $AMN$ and $AXY$ passes through the centre of the circle $ABC$. Alternatively, but equivalently, we prove that the three circles share a point different from $A$.

Invert the whole configuration from $A$ and let $Z'$ denote the image of $Z$ under inversion: The circles $ABC$, $AMN$ and $AXY$ are transformed into the lines $B'C'$, $M'N'$ and $X'Y'$, respectively; the line $BC$ is transformed into the circle $\gamma$ through $A$, $B'$ and $C'$; the circle $\gamma_B$ is transformed into a circle $\gamma'_B$ through $B'$ and $M'$, centred at $X'$ and externally tangent to $\gamma$ at $B'$; and the circle $\gamma_C$ is transformed into a circle $\gamma'_C$ through $C'$ and $N'$, centred at $Y'$ and externally tangent to $\gamma$ at $C'$. In this setting, we are to prove that the lines $B'C'$, $M'N'$ and $X'Y'$ are (projectively) concurrent.

To this end, let the pair of external common tangents of $\gamma'_B$ and $\gamma'_C$ meet at $O$, and let $\theta$ be the homothety centred at $O$ mapping $\gamma'_B$ onto $\gamma'_C$; clearly, $O$ lies on the line $X'Y'$ through the centres of the two circles. Let further $\theta_B$ and $\theta_C$ be the homotheties centred at $B'$ and $C'$, respectively, mapping $\gamma$ onto $\gamma'_B$ and $\gamma'_C$, respectively.

The centres of the three homotheties are collinear, so $B'C'$ passes through $O$.

Finally, $\theta = \theta_C\theta_B^{-1}$, so $\theta M' = \theta_C\theta_B^{-1}M' = \theta_C A = N'$, showing that the line $M'N'$ passes through $O$ as well. This ends the proof.