Question

If in a $△ABC$, $\frac{b+c}{a}=\cot \frac{A}{2}$, show that $△ABC$ is a right-angled.

Answer 1

In $△ABC,$

$\frac{b+c}{a}=\cot \frac{A}{2}$

$\frac{2R\sin B+2R\sin C}{2R\sin A}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$

$\frac{\sin B+\sin C}{2\sin \frac{A}{2}\cos \frac{A}{2}}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$

$⟹\frac{2\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{2\cos \frac{180-B-C}{2}}=\sin (90-\frac{A}{2})$

$\frac{\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{\sin \frac{B+C}{2}}=\sin \frac{B+C}{2}$

$⟹\sin (90-\frac{B-C}{2})=\sin \frac{B+C}{2}$

$⟹180-B+C=B+C⟹2B=180⟹B=90$

So $△ABC$ is a right angled triangle