What is the measure $\angle C$ from this following triangle problem?

Question

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???

Answer 1

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 $$\measuredangle C=15^{\circ}.$$

Answer 2

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 $$x=15^{\circ}.$$

Answer 3

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.