We are given a right-angled triangle $\triangle ABC$ (the right angle is $\angle ABC$).

Let $M$ be a point inside the triangle such that: $CB=AB=AM$ and $\angle BAM=30^\circ$.

Prove that $BM=CM$.

I must say that using trigonometry, specifically the cosines theorem, this problem is easy.

My goal is to find a pure geometric proof.

  • 1
    $\begingroup$ Oversimplifying, trigonometry is just giving a name ($\sin,\cos$) to some ratios. According to this point of view, everything that can be done through trigonometry can also be done without trigonometry. Just use the same ratios, without calling them $\sin\theta$ or $\cos\varphi$. $\endgroup$ – Jack D'Aurizio Nov 13 '17 at 15:07

Let $N$ be inside of the triangle $ABC$ so that triangle $BMN$ is equilateral. Then $NBA \cong MBC$ (sas) so $$CM = NA = MN = BN = MB$$


The hint:

Let $ABCD$ be a square.

Thus, $$AM=AD=MD$$ and $$\Delta BAM\cong \Delta CDM.$$


Since $AB=AM$, $\angle AMB = \angle ABM = 75^\circ$.

So we have $\angle MBC=90^\circ-75^\circ=15^\circ$.

Now we have four unknown angles. Let $ \alpha = \angle BCM$, $ \beta = \angle AMC$, $ \gamma = \angle BMC$, and $ \delta = \angle ACM$.

We can form four equations \begin{align} &\alpha+\gamma=180^\circ-15^\circ=165^\circ \\ & \alpha+\delta=45^\circ \\ & \beta+\delta=180^\circ-15^\circ=165^\circ \\ & \beta+\gamma=360^\circ-75^\circ=285^\circ \\ \end{align}

Solving, we have $ \alpha = 15^\circ =\angle MBC$. Using the property of an isosceles traingle, we have $BM=CM$.

(I can't post any image yet but here it is for reference: https://i.stack.imgur.com/k7B2N.png)


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