Finding angle in triangle I have a triangle $\triangle ABC$ with vertices $A$, $B$ and $C$ with $\angle BAC = 45^\circ$ and $\angle ABC = 30^\circ$.  Point $M$ is the center of the $BC$. Find $\angle AMC$.
This is not my homework, it's a little girl's, so she can't use trigonometric equations or anything after special lines in triangles.
Thanks in advance!
 A: Let $\measuredangle AMC=x$.
Hence, $$\frac{MC}{\sin(75^{\circ}-x)}=\frac{AC}{\sin{x}}$$ and
$$\frac{BM}{\sin(x-30^{\circ})}=\frac{AB}{\sin{x}},$$ 
which gives $$\frac{\sin(x-30^{\circ})}{\sin(75^{\circ}-x)}=\frac{\sin30^{\circ}}{\sin105^{\circ}}$$ or
$$\sin{x}\cos30^{\circ}-\frac{1}{2}\cos{x}=2\sin15^{\circ}(\sin75^{\circ}\cos{x}-\cos75^{\circ}\sin{x})$$ or
$$\tan{x}=\frac{2\sin15^{\circ}\sin75^{\circ}+\frac{1}{2}}{\cos30^{\circ}+2\sin15^{\circ}\cos75^{\circ}}$$ or
$$tan{x}=\frac{\cos60^{\circ}-\cos90^{\circ}+\frac{1}{2}}{\cos30^{\circ}+\sin90^{\circ}-\sin60^{\circ}}$$ or
$$\tan{x}=1,$$ which gives
$$x=45^{\circ}.$$
A: 
Let $X$ on $AB$ be the foot of the altitude through $C$.


*

*Since $\angle BAC = 45^\circ$, $\angle ACX = 45^\circ$, $\triangle AXC$ is isoceles and $AX = CX$.  

*Since $\angle CBA = 30^\circ$, $\angle XCB = 60^\circ$, $\triangle XBC$ is half of an equilateral triangle and $CX : BC  = 1 : 2$.

*Since $M$ is midpoint of $BC$, $CX = CM$ and $\triangle XCM$ is equilateral.

*This implies $XM = CM = BM$, $\triangle XBM$ is isosceles and $\angle BXM = \angle MBX = 30^\circ$.

*Since $AX = CX = XM$, $\triangle AXM$ is also isosceles with $\angle XAM = \angle AMX = \frac12 \angle BXM = 15^\circ$.
From this, we find
$$\begin{align}
\angle CMA &= 180^\circ - \angle AMX - \angle XMB \\
&= 180^\circ - \angle AMX - (180^\circ -\angle BXM - \angle MBX )\\
&= \angle BXM + \angle MBX - \angle AMX\\
&= 30^\circ + 30^\circ - 15^\circ\\
&= 45^\circ\end{align}$$
A: Let $A_1$ be on circumcircle of $\Delta ABC$, such that $A_1$ placed on an arc $BAC$ and $\measuredangle MA_1C=30^{\circ}$. 
Let $BK$ be an altitude of $\Delta A_1BC$ and $MK$ is an altitude of $\Delta MA_1C$.
Since $M$ is a midpoint of $BC$, we obtain $MN=\frac{1}{2}BK$ and since $\measuredangle MA_1C=30^{\circ}$, 
we see that $MN=\frac{1}{2}A_1M$.
Thus, $A_1M=BK$. 
But $\measuredangle BA_1C=\measuredangle BAC=45^{\circ}$,
which gives $BK=A_1K$ and from here $A_1M=A_1K$,
which says that $\measuredangle MKA_1=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}$.
But $MK=MB=MC$, which says that $\measuredangle BCK=75^{\circ}$, which gives $\measuredangle BCA_1=105^{\circ}$.
But we know that $\measuredangle BCA=105^{\circ}$, which gives $A\equiv A_1$.
Thus, $\measuredangle BAM=15^{\circ}$ and $$\measuredangle AMC=\measuredangle ABM+\measuredangle BAM=30^{\circ}+15^{\circ}=45^{\circ}$$
and we are done!
