The inverse of the inscribed angle theorem 
Given a triangle $\triangle ABM$ such that $|AM|=|BM|$ and a point $C$ such that the oriented angle $\angle AMB$ has twice the size of $\angle ACB$, show that $|CM|=|AM|$.

I am pretty sure that this must hold. Can somebody point me to an (elementary) proof?
 A: The circle around $M$ through $A$ (and $B$) intersects $BC$ in ($B$ and) $C'$. By the inscribed angle theorem, $2\angle AC'B=\angle AMB=2\angle ACB$, hence $AC'||AC$ (even though $\angle AC'B=\angle ACB+180^o$ is not excluded), hence $AC'=AC$ as line and $C'=C$ as intersection of that line with $BC$.
A: The answer is no.
The inscribed angle theorem ensures that for any $C$ on the red arc below, $2\angle ACB=\angle AMB$. For those points, we have that $|CM|=|AM|$. However, by symmetry, for any $C$ on the green arc below, we also have $2\angle ACB=\angle AMB$, but for $C$ on that arc $|CM|>|AM|$ (except for $A$ and $B$).
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A: The answer is no. There are two positions for $C$.
If $M$ and $C$ are on the same side of $AB$, $\angle ACM = \frac{1}{2}\angle C$ and $\angle CMA = \pi - \frac{1}{2} \angle AMB = \pi - \angle C$ and finally $\angle CAM = \pi - \angle ACM - \angle CMA = \frac{1}{2}\angle C$ and thus $\triangle AMC$ is isosceles and $CM = AM$ follows.
However, $C$ and $M$ could be on opposite sides of $AB$ (so that $AMBC$ is shaped like a kite). Then $CM = AM + 2MH$ where $MH$ is the height of $\triangle AMB$ as seen by folding $AMBC$ along $BC$ to get the first case.
