<s>[INST] Geometry Problem Isosceles Triangle 
Given this isosceles triangle, find angle AMC.
 [/INST] By using Ceva's theorem
$$\frac{AE}{EB}  \cdot \frac{BQ}{QC} \cdot \frac{CP}{PA} = 1$$
$$\frac{\sin\angle ACE}{\sin\angle ECB}\cdot \frac{\sin 10^\circ}{\sin 40^\circ}
\cdot \frac{\sin 20^\circ}{\sin 30^\circ}=1$$
$$\frac{\sin(80^\circ-\angle ACE)}{\sin\angle ACE}=\frac{\sin 10^\circ}{\sin 40^\circ}\cdot \frac{\sin 20^\circ}{\sin 30^\circ}=
\frac{\sin 10^\circ}{2\sin 20^\circ\cos 20^\circ}\cdot \frac{\sin 20^\circ}{\frac{1}{2}}=
\frac{\sin 10^\circ}{\cos 20^\circ}$$
$$\sin 80^\circ\cot\angle ACE - \cos 80^\circ = \frac{\sin 10^\circ}{\cos 20^\circ}$$
Hence we find $\angle ACE =70^\circ$ and $\angle AMC = 180^\circ - 70^\circ - 40^\circ = 70^\circ$
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