# 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|.

Given a triangle $\mathrm{△}ABM$ 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?
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Kailey Santana
Step 1
The circle around M through A (and B) intersects BC in (B and) C′. By the inscribed angle theorem, $2\mathrm{\angle }A{C}^{\prime }B=\mathrm{\angle }AMB=2\mathrm{\angle }ACB$,
Step 2
Hence $A{C}^{\prime }||AC$ (even though $\mathrm{\angle }A{C}^{\prime }B=\mathrm{\angle }ACB+{180}^{o}$ is not excluded), hence $A{C}^{\prime }=AC$ as line and ${C}^{\prime }=C$ as intersection of that line with BC.

Wevybrearttexcl
Step 1
The answer is no. There are two positions for C.
If M and C are on the same side of AB, $\mathrm{\angle }ACM=\frac{1}{2}\mathrm{\angle }C$ and $\mathrm{\angle }CMA=\pi -\frac{1}{2}\mathrm{\angle }AMB=\pi -\mathrm{\angle }C$ and finally $\mathrm{\angle }CAM=\pi -\mathrm{\angle }ACM-\mathrm{\angle }CMA=\frac{1}{2}\mathrm{\angle }C$ and thus $\mathrm{△}AMC$ is isosceles and $CM=AM$ follows.
Step 2
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 $\mathrm{△}AMB$ as seen by folding AMBC along BC to get the first case.