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 $\triangle ABC$

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 $= 11 : 4$.