Let $ \frac {\angle BNM}{\angle MNC} = k$ and $ \angle NMA + \angle MNC = h$, then we have $ 180 - k \cdot h = 2 (180 - h - 90)$ then $ k=2$, so $ \angle BNM = 2 \angle MNC$. Now let I the incenter of ABC and J of BMN. Then by angle J is the symmetric of I wrt MN and B,J,I are collinear, then $ NB = MB \ \Longleftrightarrow \ \frac{ca}{a+b} = \frac{ca}{b+c} \ \Longleftrightarrow \ c=a$.

Let $ \angle BNM=\alpha$, $ \angle MNC=k \alpha$, $ \angle BMN=\beta$, $ \angle NMA=k \beta$. Let also $ I$ be the incircle of $ \triangle ABC$. Then $ \angle AIC = \angle NIM = 180^{\circ} - \angle MNC - \angle NMA = 180^{\circ} - k \alpha - k \beta$ and also $ \angle MAC = \angle BAM = 180^{\circ} - \angle ABC - \angle BMA = 180^{\circ} - (180^{\circ} - \alpha - \beta) - (\beta + k \beta) = \alpha - k \beta$ and similarly $ \angle ACN = \beta - k \alpha$. We know that $ \angle AIC + \angle ACN + \angle CAM = 180^{\circ}$, thus $ 180^{\circ} - k \alpha - k \beta + \alpha - k \beta + \beta - k \alpha = 180^{\circ}$ which gives us $ \alpha + \beta = 2k(\alpha + \beta)$ $ 2k=1$ $ k=\frac{1}{2}$ Now draw $ \angle BNM$ and $ \angle BMN$ bisectors $ NX$ and $ NY$ respectively. Let $ O$ be the incircle of the $ \triangle NBM$ (also the $ NX$ and $ NY$ common point). Now we see that $ \angle ONM = \frac{1}{2} \angle BNM = k \alpha = \frac{1}{2} \alpha = \angle MNI$. Similarly we obtain that $ \angle OMN = \angle IMN$. Thus $ INOM$ is a deltoid and its diagonals are perpendicular, i.e. $ BI\bot NM$. But it means that $ \triangle NBM$ is isosceles ($ NB=MB$) because $ BI$ is both altitude and bisector at the same time. Thus $ \alpha = \beta$. Now we show that $ \angle BAC = 2\alpha - 2k \beta = 2\alpha - \frac{1}{2} \cdot 2\alpha = \alpha$ and similarly $ \angle ACB = \alpha$, which means that $ \angle BAC = \angle BCA$, q.e.d.