# 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?

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,

$$\angle BMC = \angle CHB = 180^\circ - \angle CBH - \angle BCH = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$

• Thank you for the response! Can you think of a solution that does not use trigonometry? Can we say that since $AC=BC$ and $AH=BH$, $CH$ is the bisector of $AB$? And we know that $\triangle ABC$ is isosceles and $AB$ is the base. Therefore, $\angle BCH=\angle ACH=\dfrac{1}{2}\angle ACB=20^\circ$. – Knowledge Greedy Mar 21 '20 at 7:35
• Also, I would like to ask how do we conclude $\angle HBM=\angle BHM=\angle HBC=10^\circ$ using the congruent triangles $BAM$ and $HAM$? – Knowledge Greedy Mar 21 '20 at 7:36
• Hey, did you see my comments? – Knowledge Greedy Mar 21 '20 at 13:08
• @Justdoit - to claim CH is a bisector, you need to show ∠BCH=∠ACH. – Quanto Mar 21 '20 at 14:50
• @Justdoit - congruent triangles BAM and HAM leads to ∠AHM=∠ABM=50, which leads to ∠HBM=∠BHM=∠HBC=10 – Quanto Mar 21 '20 at 14:52

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 $$\angle BMC=150^{\circ}$$.