Question

The angle bisectors $BD$ and $CE$ of a $\Delta ABC$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively. Then, the ratio in which $I$ divides the angle bisector through $A$ is

• A. $3:1$
• B. $11:4$
• C. $6:5$
• D. $7:4$

Answer 1

Answer: (B)

Given,
In $△ABC$

The angle bisector $BD$ and $CE$ are divided by incentre $I$ in the ratio $3:2$ and $2:1$ respectively.
$\therefore \frac{BI}{ID}=\frac{3}{2}$
and $\frac{CI}{IC}=\frac{2}{1}$
We know, $\frac{AI}{IF}=\frac{b+c}{a},\frac{BI}{ID}=\frac{a+c}{b}$,
$\frac{CI}{IE}=\frac{a+b}{c}$
$\because \frac{BI}{ID}=\frac{a+c}{b}=\frac{3}{2}$
$\Rightarrow 2(a+c)=3b...(i)$
and $\frac{CI}{IE}=\frac{a+b}{c}=\frac{2}{1}$
$\Rightarrow a+b=2c...(ii)$
On solving Eqs. (i) and (ii), we get
$b=\frac{3}{2}a$ and $c=\frac{5}{4}a$
$\because \frac{AI}{IF}=\frac{b+c}{a}=\frac{\frac{3}{2}a+\frac{5}{4}a}{a}$
$=\frac{11}{4}$
Hence, ratio $=11:4$.