Question

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

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

Answer 1

The correct option is B $11:4$

Incentre divides the angle bisector in a defined ratio of the length of sides of the triangle

$\frac{AI}{IF}=\frac{b+c}{a}......\left(i\right)\frac{BI}{ID}=\frac{a+c}{b}=\frac{3}{2}\Rightarrow 2a+2c=3b......\left(ii\right)\frac{CI}{IB}=\frac{a+b}{c}=\frac{2}{1}\Rightarrow 2c=a+b.....\left(iii\right)$

using $\left(iii\right)$ in $\left(ii\right)$

$2a+a+b=3b3a=2bb=\frac{3a}{2}....\left(iv\right)$

Again $2c=a+b$

$2c=a+\frac{3a}{2}c=\frac{5a}{4}.....\left(v\right)$

using $\left(iv\right)$ and $\left(v\right)$ in $\left(i\right)$

$\frac{AI}{IF}=\frac{\frac{3a}{2}+\frac{5a}{4}}{a}=\frac{22}{8}=\frac{11}{4}$

So option $B$ is correct