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?
Isosceles triangle $ABC$ with an inside point $M$, find $\angle BMC$
geometry
Related Solutions
I saw the following solution may years ago:
On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.
Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.
But then $$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$
Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.
Let $E$ be a point inside the triangle such that $EB=EC=BC$.
$\qquad\qquad\qquad$
Since $\measuredangle{ABE}=\frac{180^\circ-20^\circ}{2}-60^\circ=20^\circ$, we can see that $\triangle{ABD}$ and $\triangle{BAE}$ are congruent.
Hence, $$\measuredangle{DBC}=\measuredangle{ABC}-\measuredangle{ABD}=\measuredangle{ABC}-\measuredangle{BAE}=80^\circ-10^\circ=\color{red}{70^\circ}.$$
Best Answer
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,
$$\angle BMC = \angle CHB = 180^\circ - \angle CBH - \angle BCH = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$