What is the measure $\angle C$ from this following triangle problem? For reference:  In a triangle ABC a $m \measuredangle A = 2m \measuredangle C$.
Plot the median BM such that $\measuredangle AMB=45º$. Calculate the $m \measuredangle C$.
My progress:
$Trace DM \perp AC\\
EXtend~AB~ to~D\\
\text{mark point E (circuncentro) on the straight DM in the }  \triangle ADC\\\overset{\LARGE{\frown}}{CD}=2x \therefore \measuredangle AEC = 4x $

it remains to demonstrate that EAM =2x???
 A: By law of sines we obtain:
$$\frac{\sin2x}{\sin(135^{\circ}-2x)}=\frac{BM}{AM}=\frac{BM}{CM}=\frac{\sin x}{\sin(45^{\circ}-x)},$$ which gives
$$2\cos x(\cos x-\sin x)=\sin2x+\cos2x$$ or
$$1+\cos2x-\sin2x=\sin2x+\cos2x$$ or
$$\sin2x=\frac{1}{2},$$ which gives $$x=15^{\circ}.$$
A: Another trigonometric approach: Using cot theorem, $$2\cot 45^\circ=\cot x-\cot 2x$$ $$\implies 2=\frac 1t-\frac{1-t^2}{2t}\tag{t=tan x}$$ $$\implies t^2-4t+1=0$$ $$\implies t=(2-\sqrt 3) \ \text{or} \ t=(2+\sqrt 3)$$ though $t\lt 1$, therefore, $$\tan x=(2-\sqrt 3)\implies x=15^\circ$$ P.S. If you don't remember $\tan 15^\circ=(2-\sqrt 3)$, you can convert $\cot$ into $\sin$ and $\cos$ in the first equation. This will lead you to $\sin 2x=\frac 12$ which you should remember.
A: Another way.
Let $G$ be a mid-point of $AB$ and $BE$ be an altitude of $\Delta ABC$.
Thus, since $AG=GE=GB,$ $BE=EM$ and
$GE=EM$ (because $GM||BC$, $\measuredangle GMA=\measuredangle C$ and $\measuredangle GEA=\measuredangle A=2\measuredangle C$ ), we obtain:
$$EB=EM=EG=\frac{1}{2}AB,$$ which gives $$\measuredangle BAC=30^{\circ}$$ and $$\measuredangle C=15^{\circ}.$$
