Question

The angle bisectors BD and CE of a triangle ABC are divided by the incentre I in the ratio of 3 : 2 and 2 :1 respectively. Then the ratio in which I divide the angle bisector through A is.

• A. 3:1
• B. 11:4
• C. 6:5
• D. 7:4

Answer 1

The correct option is B

11:4

$\therefore \frac{AI}{IF}=\frac{b+c}{a}.........\left(1\right)$

$\because \frac{BI}{ID}=\frac{a+c}{b}=\frac{3}{2}\left(2\right)$

$\because \frac{CI}{IE}=\frac{a+b}{c}=\frac{2}{1}$

$\Rightarrow a+b=2c\left(3\right)$

$\Rightarrow 2a+2c=3b$

$\Rightarrow 2a+a+b=3busing\left(3\right)$

$\Rightarrow 3a=2b$

$\Rightarrow b=\frac{3}{2}a....\left(4\right)$

Now~again~(3)$\Rightarrow 2c=a+b$

$2c=a+\frac{3}{2}a$

$\Rightarrow c=\frac{5}{4}a$

Hence $\frac{AI}{IF}=\frac{b+c}{a}=\frac{\frac{3}{2}a+\frac{5}{4}a}{a}=\frac{11}{4}$