Question 45
In triangle $ABC$ the points $M$ and $N$ are the midpoints of the sides $AB$ and $AC$, respectively. The lines $BN$ and $CM$ intersect the circumcircle of $ABC$ for the second time at $N'$ and $M'$, respectively. Points $X$ and $Y$ are on the extension of $BC$ such that point $B$ is between $X$ and $C$ and point $C$ is between $B$ and $Y$, and
$$
\angle BXM' = \angle ACM, \quad \angle CYN' = \angle ABN
$$
Prove that $AY = AX$.
$$
\angle BXM' = \angle ACM, \quad \angle CYN' = \angle ABN
$$
Prove that $AY = AX$.