<s>[INST] 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!
 [/INST] 
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
$$\begin{align}
\angle CMA &= 180^\circ - \angle AMX - \angle XMB \\
&= 180^\circ - \angle AMX - (180^\circ -\angle BXM - \angle MBX )\\
&= \angle BXM + \angle MBX - \angle AMX\\
&= 30^\circ + 30^\circ - 15^\circ\\
&= 45^\circ\end{align}$$
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