find the measure of $AMC$ if $M$ is the midpoint of $BC$ then find the measure of $AMC$.

I tried to use the angles to find $AMC$ but I don't know how to use that $M$ is the midpoint of $BC$.
 A: I asked this from our teacher and he give this answer may helps also this answer don't uses trigonometry:

$BNC=90$degrees and $BAN=45$ then we have $AN=BN$ also because $BNC$ is right angle $MN=BM$ also because $NBM=60$,the triangle $BNM$ is equilateral.Also $MNA=30$ then because $MN=AN$ the angle $NMA=75$ then we have $AMC=45$.
A: The answer is $\color{red}{45^\circ}$. You may embed the construction in the complex plane by assuming $C=1, B=-1$. Then $A$ has to fulfill:
$$A-C = \lambda e^{5\pi i/6},\qquad A-B = \mu e^{\pi i/12},\qquad (\lambda,\mu\in\mathbb{R}) $$
but that leads to $A=\eta(1+i)$, and the claim follows.

An alternative, elementary way. By assuming $BM=MC=1$ we have:
$$ AC = 2\frac{\sin 15^\circ}{\sin 135^\circ}=-1+\sqrt{3},\qquad AB = 2\frac{\sin 30^\circ}{\sin 135^\circ}=\sqrt{2}$$
by the sine theorem. By Stewart's theorem we have:
$$ AM^2 = \frac{2AB^2+2AC^2-BC^2}{4} = \left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)^2$$
and by the sine theorem again:
$$ \sin\widehat{CMA} = AC\cdot\frac{\sin 30^\circ}{AM} = \frac{1}{\sqrt{2}},$$
from which $\widehat{CMA}=\color{red}{45^\circ}$.
A: Let $X$ be the point on $BC$ exactly below $A$ and $MX=z$ for comfort.
$BX=\frac{BC}{2}+z$ and $XC=\frac{BC}{2}-z$
Now:
$$\tan(\alpha )= \frac{AX}{BX}\implies \tan(\alpha )= \frac{AX}{\frac{BC}{2}+z}$$
$$\tan(AMC )= \frac{AX}{Z}$$
$$\tan(\beta )= \frac{AX}{XC} \implies \tan(\beta )= \frac{AX}{\frac{BC}{2}-z}$$
Now you have 3 questions with 3 variables $z,AX,\tan(AMC)$ and the rest is given. Let $AMC= \gamma$
$$AX=z \tan(\gamma)$$
$$(1)\tan(\alpha )= \frac{z \tan(\gamma)}{\frac{BC}{2}+z}\implies \frac{BC\tan(\alpha )}{2}=z(\tan(\gamma)-\tan(\alpha ))$$
$$(2)\tan(\beta )= \frac{z \tan(\gamma)}{\frac{BC}{2}+z}\implies \frac{BC\tan(\beta )}{2}=z(\tan(\gamma)+\tan(\beta ))$$
And by dividing (1)/(2) we get:
$$\frac{\tan(\alpha )}{\tan(\beta )}=\frac{\tan(\gamma)-\tan(\alpha )}{\tan(\gamma)+\tan(\beta )}$$
Hence:
$$\tan(\gamma)=\frac{2\tan(\alpha)\tan(\beta)}{\tan(\beta)-\tan(\alpha)}$$
$$\gamma=\arctan(\frac{2\tan(30^\circ)\tan(15^\circ)}{\tan(30^\circ)-\tan(15^\circ)})$$
A: 
Continue $CA$, draw the ray $\vec{CA}$. Then draw $BN$, in which it is perpendicular to $\vec{CA}$. 
$\triangle BCN$ is a 30-60-90 triangle. as we know about this triangle. $BN=\frac12 BC$. So $BN=BM$. This tells us $\triangle ABN$, is isosceles, so $\angle BMN=\angle BNM=\frac{180^\circ-\angle MBN}2=\frac{120^\circ}{2}=60^\circ$. 


*

*Now we know $\triangle BMN$ is equilateral.


Therefor $$MN=BN=BM\qquad(1)\\
\text{and }\angle MBN=60^\circ\qquad{ }$$
By this, we get $\angle CMN=180^\circ-\angle BMN=180^\circ-60^\circ$, 
$$\text{therefor }\angle CMN=120^\circ\  (2)$$
$\angle ABN=\angle CBN-\angle CBA=60^\circ-15^\circ=45^\circ$. So, the other angle of the right triangle $\triangle ABN$, i.e. $\angle BAN$, should be $180^\circ-90^\circ-45^\circ=45^\circ{}^\dagger$, therefor $\triangle ABN$ is isosceles.
So we would have $AN=BN$. By (1) we get $MN=AN$, so $\triangle AMN$ is isosceles. Therefor $$\angle AMN=\angle MAN=\frac{180^\circ-\angle ANM}2=\frac{180^\circ-30^\circ}2=75^\circ$$
Now, by using (2) we can find the size of $\angle AMC$, which is 
$$\angle AMC=\angle CMN-\angle AMN=120^\circ-75^\circ=45^\circ$$
${}^\dagger$let's note that one may also know that $\angle BAN$ is equal to $45^\circ$, since it is an external angle of the triangle  $\triangle ABC$
A: Another geometric solution:

Let $E$ be the foot of the altitude through $A$. Let $D$ be the midpoint of $\overline{AC}$, so $\triangle EDC$ is isosceles with $\angle DEC=\angle ECD=30$ and $ED = DC$.
Both $M$ and $D$ are midpoints, so $\triangle ABC\sim\triangle DMC$. Conclude $\angle DMC=\angle ABC=15$. Knowing $\angle DEC=30$, compute $\angle EDM=15$ so $\triangle MED$ is isosceles. Hence $ME=ED$.
Since $D$ is the midpoint of $AC$, we know $DC=\frac12 AC$. But $\triangle AEC$ is a 30-60-90 triangle so also $AE=\frac12 AC$.
Putting it all together, we've shown
$ME=ED=DC=\frac12 AC=AE$. So $\triangle MEA$ is isosceles. Since $\angle MEA$ is right, it follows that $\angle AME=45$.
