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:

enter image description here

$BN/BC = 2BM/MA$

$1+\frac{NC}{BN} = \left(\frac{MA}{2BM}+\frac{1}{2}\right) + \frac{1}{2}$

$2\frac{BC}{BN}= \frac{AB}{AM}+1$

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!

  • $\begingroup$ Thank you. Please explain why, $\frac{2BM}{MC}=\frac{2S_{\Delta BMN}}{S_{\Delta AMN}}$. I understand the rest. $\endgroup$ – carat Jul 23 '17 at 7:22
  • $\begingroup$ @carat It was typo. See now please. $\endgroup$ – Michael Rozenberg Jul 23 '17 at 7:46
  • $\begingroup$ I'm clear now. Thank you for your kind help. $\endgroup$ – carat Jul 23 '17 at 8:20
  • $\begingroup$ @carat You are welcome! $\endgroup$ – Michael Rozenberg Jul 23 '17 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.