Geometry Problem Isosceles Triangle 
Given this isosceles triangle, find angle AMC.
 A: 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 $\angle AMC = 180^\circ - 70^\circ - 40^\circ = 70^\circ$
A: You know that the angles of a triangle add up to $180°$. Use that fact to calculate angle $AQC$. Then with this, calculate angle $BQM$ and then $QMB$. Continue to calculate $BPC$ to therefore obtain $BPA$. Continue with this process till you find the desired angle.
A: To provide a geometric argument, my construction goes as follows.
Draw a line AR below AB, such that the length of AR (that I denote m(AR), the measure of AR) is equal to m(AC). Thus m(AR) = m(AC). Do that such that angle RAB is 10 degrees. Connect C and R. But m(AC) = m(AR) and angle RAC is 60 degrees. Thus triangle CAR is equilateral. In particular, angle ACR is 60 degrees. Thus angle RCB is 20 degrees and triangle RCB is isosceles since m(CR)=m(CB). Thus angle ABR is 30 degrees (completing the angles in the isosceles triangle RCB).
Remark that triangles AMB and ARB are congruent (they have a common side AB, and interior angles are 10-140-30 in both cases). Thus m(MB) = m(RB), but angle RBM is 60 degrees. Thus triangle RBM is equilateral. Triangle RMB is equilateral, in the isosceles triangle RCB. Thus line CM divides angle RCB into two equal angles. Each is 10 degrees. The sum of the angles in triangle BMC is 180 degrees.
Given that angle MCB is 10 degrees and angle MBC is 20 degrees, we conclude that angle CMB is 150 degrees, and the desired angle is 360 - 140 - 150 = 70 degrees.
