Question

If x,y and z are the distances of incenter from the vertices of the triangle ABC, respectively, then prove that $\frac{abc}{xyz}=$

• A. $\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
• B. $\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$
• C. $\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
• D. $\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

Answer 1

The correct option is A $\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$

$x=rcosec\frac{A}{2}$ and $a=r(\cot \frac{B}{2}+\cot \frac{C}{2})$ $\Rightarrow \frac{a}{x}=(\cot \frac{B}{2}+\cot \frac{C}{2})\sin \frac{A}{2}=\frac{\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}$ or
$\frac{abc}{xyz}=\frac{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}=\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$