Question

If $p,q,r$ are the length of the internal bisector of the angles, $A,B,C$ of a $△ABC$, respectively then $\frac{1}{p}\cos \frac{A}{2}+\frac{1}{q}\cos \frac{B}{2}+\frac{1}{r}\cos \frac{C}{2}$

• A. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
• B. $\frac{1}{a}+\frac{1}{c}-\frac{1}{b}$
• C. $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}$
• D. $\frac{1}{c}+\frac{1}{b}-\frac{1}{a}$

Answer 1

The correct option is A $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

$\Delta =\frac{1}{2}pc\sin \frac{A}{2}+\frac{1}{2}pb\sin \frac{A}{2}$

Also, $\Delta =\frac{1}{2}bc\sin A=bc\sin \frac{A}{2}\cos \frac{A}{2}$

$\therefore bc\sin \frac{A}{2}\cos \frac{A}{2}=\frac{1}{2}pc\sin \frac{A}{2}+\frac{1}{2}pb\sin \frac{A}{2}$

$\Rightarrow \frac{1}{p}\cos \frac{A}{2}=\frac{1}{2}(\frac{1}{b}+\frac{1}{c})$

$\therefore \frac{1}{p}\cos \frac{A}{2}+\frac{1}{q}\cos \frac{B}{2}+\frac{1}{r}\cos \frac{C}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$