Isosceles triangle $ABC$ with an inside point $M$, find $\angle BMC$ We have an isosceles $\triangle ABC, AC=BC, \measuredangle ACB=40^\circ$ and a point $M$ such that $\measuredangle MAB=30^\circ$, $\measuredangle MBA=50^\circ$.
Find $\measuredangle BMC$.

Starting with $\angle ABC=\angle BAC=70^\circ \Rightarrow \angle CBM=20 ^\circ$. Let us construct the equilateral $\triangle ABH$. If we look at $\triangle ACH, \angle ACH=20^\circ$ and $\angle CAH=10^\circ$. Can we show $\triangle AHC \cong CHB$? Any other ideas?
 A: 
Construct the equilateral triangle $AHB$. Given that $AC = BC, AH = BH$ and the shared $CH$, the triangles $AHC$ and $BHC$ are congruent. Then, $\angle BCH = \dfrac12\angle ACB = 20^\circ$.
Since $AH = BH$ and $\angle BAM = \angle HAM = 30^\circ$, the triangles $BAM$ and $HAM$ are congruent, which yields $\angle HBM = \angle BHM = \angle HBC = 10^\circ$ and $HM || CB$. 
Then, the triangles $CHB$ and $BHC$ have the same altitudes $h$ with respect to the base $BC$. Since $\angle BCH = \angle CBM = 20^\circ$, we have $CH = BM = h\cot 20^\circ$.
As a result, the triangles $CHB$ and $BMC$ are congruent, which leads to,
$$\angle BMC = \angle CHB = 180^\circ - \angle CBH - \angle BCH = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$
A: Here's a trigonometric approach. Let $\angle BCM=\varphi\Rightarrow \angle ACM=40^{\circ}-\varphi$. Apply the law of sines in $\triangle AMC$ and $\triangle BMC$:
$$\frac{AC}{CM}=\frac{\sin(80^{\circ}-\varphi)}{\sin(40^\circ)} \\ \frac{BC}{CM}=\frac{\sin(20^{\circ}+\varphi)}{\sin(20^\circ)} $$
Since $AC=BC$, the two ratios with the sines are equal. We have $\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)$, so
$$\frac{\sin(80^{\circ}-\varphi)}{2\cos(20^\circ)}=\sin(20^\circ+\varphi) \Leftrightarrow \\ \sin(80^{\circ}-\varphi)=2\sin(20^\circ+\varphi)\cos(20^\circ)$$ Then use the sum-product identities:
$$\sin(80^{\circ}-\varphi)=\sin(\varphi)+\sin(\varphi+40^\circ) \Leftrightarrow \\
\sin(\varphi)=\sin(80^{\circ}-\varphi)-\sin(\varphi+40^\circ) \Leftrightarrow \\
\sin(\varphi)=2\sin(20^\circ-\varphi)\cos(60^\circ) \Leftrightarrow \\
\sin(\varphi)=\sin(20^\circ-\varphi) $$
Since $0<\varphi<40^{\circ}$, the last equality implies $\varphi=20^\circ-\varphi\Leftrightarrow \varphi=10^{\circ}$, and we find $\angle BMC=150^{\circ}$.
