The angle bisectors BD and CE of a triangle A | vedclass

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(B) Given,in $	riangle ABC$,the angle bisectors $BD$ and $CE$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively.$\frac{BI}{I

Vedclass · Accepted Answer

$11:4$

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(B) Given,in $	riangle ABC$,the angle bisectors $BD$ and $CE$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively.$\frac{BI}{ID} = \frac{3}{2}$ and $\frac{CI}{IE} = \frac{2}{1}$.We know that for an incentre $I$ of $	riangle ABC$ with sides $a, b, c$ opposite to vertices $A, B, C$ respectively,the ratio in which $I$ divides the angle bisectors are:$\frac{AI}{IF} = \frac{b+c}{a}$,$\frac{BI}{ID} = \frac{a+c}{b}$,and $\frac{CI}{IE} = \frac{a+b}{c}$.Given $\frac{a+c}{b} = \frac{3}{2} \Rightarrow 2a + 2c = 3b \quad \dots(i)$Given $\frac{a+b}{c} = \frac{2}{1} \Rightarrow a + b = 2c \quad \dots(ii)$From $(ii)$,$c = \frac{a+b}{2}$. Substituting this into $(i)$:$2a + 2(\frac{a+b}{2}) = 3b$$2a + a + b = 3b$$3a = 2b \Rightarrow b = \frac{3}{2}a$Now,substitute $b = \frac{3}{2}a$ into $(ii)$:$a + \frac{3}{2}a = 2c$$\frac{5}{2}a = 2c \Rightarrow c = \frac{5}{4}a$Finally,the ratio $\frac{AI}{IF} = \frac{b+c}{a} = \frac{\frac{3}{2}a + \frac{5}{4}a}{a} = \frac{6+5}{4} = \frac{11}{4}$.Thus,the ratio is $11:4$.

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,the ratio in which $I$ divides the angle bisector through $A$ is

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