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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