We claim that $A_1 B_2$ and $A_2 B_1$ intersect at $C$. Define $O_A$ to be the midpoint of $A_1 A_2$ and $O_B$ the midpoint of $B_1 B_2$. Then we have the equalities $O_A A_1 = O_A C = O_A A_2, O_B B_1 = O_B C = O_B B_2$. Then, since $\frac{O_A A_1}{O_B B_2} = \frac{O_A C}{O_B C}, \angle A_1 O_A C = \angle B_2 O_B C$ (the second since $A_1 A_2 \parallel B_1 B_2$, we find that $\triangle A_1 O_A C \sim \triangle B_2 O_B C$. Since $A_1$ and $B_2$ are on opposite sides of $O_A O_B$, we conclude that $A_1, C, B_2$ are collinear. We can repeat the same argument with $A_2, C, B_1$. Now, since the point of intersection of lines $A_1 B_2$ and $A_2 B_1$ is simply $C$, the circle with the center on the common internal tangent to the two circles which also passes through this point is simply the point $C$. Then the points $M$ and $N$ are both also $C$, and thusly we trivially have $MN \perp A_1 A_2, B_1 B_2$ and we are done.