Question

If $\alpha ,\beta ,\gamma$ are the lengths of internal bisectors of angles $A,B,C$ respectively of $\Delta ABC$, then $\sum \frac{1}{\alpha }\cos \frac{A}{2}$ is:

• A. $a+b+c$
• B. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
• C. $2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
• D. $2(a+b+c)$

Answer 1

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

We know that,
$\alpha =\frac{2bc}{b+c}\cos \frac{A}{2}$
$\Rightarrow \frac{1}{\alpha }\cos \frac{A}{2}=\frac{b+c}{2bc}\cdots \left(i\right)$
$\beta =\frac{2ac}{a+c}\cos \frac{B}{2}$
$\Rightarrow \frac{1}{\beta }\cos \frac{B}{2}=\frac{a+c}{2ac}\cdots \left(ii\right)$
$\gamma =\frac{2ab}{a+b}\cos \frac{C}{2}$
$\Rightarrow \frac{1}{\gamma }\cos \frac{C}{2}=\frac{a+b}{2ab}\cdots \left(iii\right)$
Adding Equation $\left(i\right),\left(ii\right),\left(iii\right)$, we get
$\sum \frac{1}{\alpha }\cos \frac{A}{2}=\frac{b+c}{2bc}+\frac{a+c}{2ac}+\frac{a+b}{2ab}$
$\Rightarrow \sum \frac{1}{\alpha }\cos \frac{A}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$