an angle ratio

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Source: Mathematics Regional Olympiad of Mexico Southeast 2015 P5

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AlanLG

#1 h Oct 26, 2021, 10:14 PM • 1 Y

Y by HWenslawski

In the triangle $ABC$ , let $AM$ and $CN$ internal bisectors, with $M$ in $BC$ and $N$ in $AB$ . Prove that if

$\frac{\angle BNM}{\angle MNC}=\frac{\angle BMN}{\angle NMA}$
then $ABC$ is isosceles.

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#2 h Nov 23, 2022, 8:09 PM

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Let
$\angle BNM = \alpha, \angle MNC = \beta, \angle BMN = \theta$ and $\angle NMA = \gamma.$
Then
$\frac{\alpha}{\beta} = \frac{\theta}{\gamma} \Rightarrow \frac{\gamma}{\beta} = \frac{\theta}{\alpha} \Rightarrow \frac{\gamma + \beta}{\beta} = \frac{\theta+ \alpha}{\alpha}.$
Notice that
$\gamma + \beta = \frac{\angle A + \angle C} {2}$ and $\theta + \alpha = 180^{\circ} - \angle ABC = \angle A + \angle C.$
So $\frac{\angle A + \angle C}{2 \beta} = \frac{\angle A + \angle C}{\alpha} \Rightarrow 2 \beta = \alpha$ , that is, $\frac{\alpha}{\beta} = \frac{\theta}{\gamma} = 2.$
Now, let $I'$ be the incenter of $\triangle BNM$ and let $I$ be the incenter of $\triangle ABC$ . It's clear that $\triangle INM \sim \triangle I'NM$ and $BI \perp NM$ , thus $\triangle BNM$ is isosceles.
So $\beta = \gamma$ and the quadrilateral $I'NIM$ is a rhombus with $I'N \parallel MA$ and $I'M \parallel NC$ so that $\frac{\angle C}{2} = \gamma = \beta = \frac{\angle A}{2}$ , therefore $ABC$ is isosceles with $BA = BC$ .

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