Question

If the interior and exterior bisectors of the $\angle$A of a $\Delta$ABC meet the base $BC$ at $D$ and $E$, then

• A. $2BC=BD+BE$
• B. $B{C}^2=BD\times BE$
• C. $\frac{2}{BC}=\frac{1}{BD}+\frac{1}{BE}$
• D. None of these

Answer 1

The correct option is C $\frac{2}{BC}=\frac{1}{BD}+\frac{1}{BE}$

Let A be the initial point and the position vector of B and C be b and c, respectively. Hence, let AB $=$ c and AC $=$ b. The internal bisector is $r=t(\frac{b}{c}+\frac{c}{b})$ and the equation of the line BC is

$r=b+k(c-b)$

and hence the position vector of D is $\frac{bb+cc}{b+c}$

$\Rightarrow BD=\frac{bb+cc}{b+c}-b=\frac{c(c-b)}{b+c}$

$\Rightarrow BD=\frac{c|c-b|}{b+c}=\frac{ca}{b+c}$

Similarly, $BE=\frac{c|c-b|}{c-b}=\frac{ca}{c-b}$

and hence, $\frac{1}{BE}+\frac{1}{BD}=\frac{c-b}{ac}+\frac{c+b}{ac}$

$\frac{1}{BE}+\frac{1}{BD}=\frac{2c}{ac}=\frac{2}{a}=\frac{2}{BC}$