I have a hard time solving the following problem:
Given an Isosceles triangle $\triangle ABC$ where $AC = BC$, with $\angle ACB = 50^{\circ}$ let $M$ be a point outside the triangle $\triangle ABC$ but within the angle $\angle BAC$. If $\angle AMB = 25^{\circ}$ and $\angle AMC = 20^{\circ}$ then what is the angle $\angle BCM$.
I drew a picture which is linked below of the problem but I have a hard time knowing how to solve this one since merely using the fact that the angle sum of a triangle is $180^{\circ}$ isn't enough to solve it since we are left with one unknown.
If we let $\angle BAM = \delta$ then
\begin{align*} \angle MAC & = 65^{\circ}-\delta\\ \angle BCM & = 45^{\circ}+\delta \end{align*}
I would be very thankful for some tip on how to get started with the exercise.
We know that if we let $C$ be the center of a circle with radius $r = AC = BC$ then any point $M$ on the circumference must satisfy $\angle AMB = 25^{\circ}$ but how do we know the reversed statement that since $\angle AMB = 25^{\circ}$ then $M$ lies on the circle? Can we use that the triangles are congruent?
Since $AC = BC$ and $$\angle ACB = 2\cdot \angle AMB$$ we see that $M$ is on a circle with center at $C$ and radius $r=AC = BC$. So $CM =CB$ and so $\angle MBC = 45$ so $$\angle BCM =90$$