How can I show that $AX=AY$?

Question

In the diagram below, the two circles have equal radii. However after much bashing around with angles, I was not able to show that $AX=AY$.

My only idea so far is to try to instead prove that $\angle AXY = \angle AYX$, yet I was still unable to do this. We of course need to incorporate the fact that both circles have equal radii at some point, but I cannot see a way of doing this without lots of constructions and tedious angle chasing.

Answer 1

The small arcs $AX$ of the left circle and $AY$ of the right circle both span the same angle $\angle ABY$.

Answer 2

If the center on the left is $O_1$ and on the right is $O_2$, then $$ \angle ABY=0.5 \angle AO_1X = 0.5 \angle AO_2Y $$

Answer 3

As figure shows, $\angle AZB+\angle AYB=180^\circ.$ Moreover, since the two circle have the same size, by using the property of the inscribed angles, $\angle AZB=\angle AXB$. Therefore $\angle AXB+\angle AYB=180^\circ,$ that is, $\angle AXY=\angle AYX.$ Hence $AX=AY.$