A very interesting problem in geometry, with a surprisingly easy solution

Question

I just came across the following question:

Let the circles $k_1$ and $k_2$ intersect at two points $A$ and $B,$ and let $t$ be a common tangent of $k_1$ and $k_2$ that touches $k_1$ and $k_2$ at $M$ and $N$ respectively. If $t$ is perpendicular to $AM$ and $MN=2AM$ evaluate the angle $NMB$

I solved it in the following way:

$CB\cdot CA=(CN)^2$

$CB\cdot CA=(CM)^2$

So $CN=CM$

So $CM=MA$, $MCA=MAC=45$

Since $AMC=90$, then $BMC=45$

This is the first time I have actually managed to solve a geometry questions of this type and am surprised by its simplicity, could you please tell me if I am correct and propose alternative solutions?

Answer 1

Instead of $CM = CA$, it should be $$CM = AM.$$ Also, you did not formally establish why $\angle BMC = 45^\circ$. The reason is because $\triangle AO_1 B$ is 45-45-90 since $AO_1 = BO_1$ and $\angle O_1 A B = 45^\circ$. Hence $$\triangle MAC \sim \triangle O_1 AB$$ and $B$ is the midpoint of $AC$.