Incenter divide ratio

Question

Given a triangle $ABC$ and angle bisectors $BD,CE$ which intersect at $O$ (incenter) . The ratio in which $O$ divides $BD$ is $3:2$ and it divides $CE$ in ratio $1:2$ . Find the ratio in which the third bisector is divided by the incenter .

Answer 1

From the given, $\frac{\Delta AOB}{\Delta ABC}=\frac{2}{1+2}=\frac{2}{3}$ and $\frac{\Delta AOC}{\Delta ABC}=\frac{2}{2+3}=\frac{2}{5}$. So $\frac{\Delta AOB+\Delta AOC}{\Delta ABC}=\frac{16}{15}>1$. How can this be possible?

Answer 2

In-center divides the angle bisectors in some definite ratio. Let $ABC$ be a triangle $AB = b$, $BC = a$ , $AC = c$ & let bisector of angle $A$ = $AF$ touching $BC$ at $F$ . Then $AF$ is divided in the ratio $(b+c): a$. Proceeding in this manner for other sides you will come to your answer . The answer will be $11:4$. Thank you.

Answer 3

This question was answered by Kaustubh mishra

angle bisector theorem: the angle bisector of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

$I$ (incenter), $AB=c$, $BC=a$, $AC=b$, $AD=m$, $BD=n$, $m+n=c$

$\begin{array}{} ΔACD & \frac{CI}{ID}=\frac{b}{m}=r & ⇒ & m=\frac{b}{r} & (1) \\ ΔBCD & \frac{CI}{ID}=\frac{a}{n}=r & ⇒ & n=\frac{a}{r}& (2) \\ \end{array}$

$\begin{array}{} \text{adding (1) with (2), we get} & c=\frac{a+b}{r} & ⇒ & r=\frac{CI}{ID}=\frac{a+b}{c} \end{array}$

$\begin{array}{} r_{A}=\frac{b+c}{a} & (3) & r_{B}=\frac{a+c}{b} & (4) & r_{C}=\frac{a+b}{c} & (5) \end{array}$

$\begin{array}{} \text{isolating a in (4) and (5)} & a=b·r_{B}-c & (6) & a=c·r_{C}-b & (7) & ⇒ & c=\frac{1+r_{B}}{1+r_{C}}·b & (8) \end{array}$

$\begin{array}{} \text{replacing (8) in (6)} & a=\frac{r_{B}·r_{C}-1}{r_{C}+1}·b & (9) \end{array}$

$\begin{array}{} \text{replacing (8) and (9) in (3)} & r_{A}=\frac{r_{B}+r_{C}+2}{r_{B}·r_{C}-1} \end{array}$

$\begin{array}{} \text{for} & r_{B}=\frac{3}{2} & \text{and} & r_{C}=\frac{2}{1} & \text{we get} & r_{A}=\frac{11}{4} \end{array}$

$\begin{array}{} \text{constructing triangle ABC for c = 1} & b=\frac{6}{5}=1.2 & \text{by (8) and (9) we get} & a=\frac{4}{5}=0.8 \end{array}$