Given a triangle $\mathrm{\u25b3}ABM$ such that $|AM|=|BM|$ and a point C such that the oriented angle $\mathrm{\angle}AMB$ has twice the size of $\mathrm{\angle}ACB$, show that $|CM|=|AM|$.

I am pretty sure that this must hold. Can somebody point me to an (elementary) proof?

I am pretty sure that this must hold. Can somebody point me to an (elementary) proof?