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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 $\measuredangle BMC$. enter image description here 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?

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2 Answers 2

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enter image description here

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,

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

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  • $\begingroup$ 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, $\angle BCH=\angle ACH=\dfrac{1}{2}\angle ACB=20^\circ$. $\endgroup$ Commented Mar 21, 2020 at 7:35
  • $\begingroup$ Also, I would like to ask how do we conclude $\angle HBM=\angle BHM=\angle HBC=10^\circ$ using the congruent triangles $BAM$ and $HAM$? $\endgroup$ Commented Mar 21, 2020 at 7:36
  • $\begingroup$ Hey, did you see my comments? $\endgroup$ Commented Mar 21, 2020 at 13:08
  • $\begingroup$ @Justdoit - to claim CH is a bisector, you need to show ∠BCH=∠ACH. $\endgroup$
    – Quanto
    Commented Mar 21, 2020 at 14:50
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    $\begingroup$ @Justdoit - congruent triangles BAM and HAM leads to ∠AHM=∠ABM=50, which leads to ∠HBM=∠BHM=∠HBC=10 $\endgroup$
    – Quanto
    Commented Mar 21, 2020 at 14:52
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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 $\angle BMC=150^{\circ}$.

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