Let $ABC$ be a triangle. Points $M$ and $N$ are on the sides $AB$ and $BC$ respectively such that $\frac{BN}{BC} = \frac{2BM}{MA}$ and $\angle BNM=\angle ANC$. Prove that ABC is a right-angled triangle.
My attempted work:
$BN/BC = 2BM/MA$
$1+\frac{NC}{BN} = \left(\frac{MA}{2BM}+\frac{1}{2}\right) + \frac{1}{2}$
$2\frac{BC}{BN}= \frac{AB}{AM}+1$
I've tried to draw several auxiliary lines and still couldn't do it.
Let $\measuredangle BNM=\alpha$.
Thus, $$\frac{2BM}{MA}=\frac{2S_{\Delta BMN}}{S_{\Delta AMN}}=\frac{2BN\cdot MN\sin\alpha}{AN\cdot MN\sin(180^{\circ}-2\alpha)}=\frac{BN}{AN\cos\alpha}.$$ In another hand, $$\frac{2BM}{MA}=\frac{BN}{NC}.$$ Thus, $$NC=AN\cos\alpha.$$ Let $\measuredangle C=\gamma$.
Thus, by law of sines for $\Delta ANC$ we obtain: $$\cos\alpha=\frac{\sin(\alpha+\gamma)}{\sin\gamma}$$ or $$\cos\alpha\sin\gamma=\sin\alpha\cos\gamma+\cos\alpha\sin\gamma,$$ which gives $\gamma=90^{\circ}$ and we are done!