Question

In $\Delta ABC,AD$ is the bisector of $\angle BAC$ and $I$ is its incentre .

Prove that : $\frac{AI}{ID}=\frac{AB+AC}{BC}$.

Answer 1

Given:

In $\Delta ABC,AD$ is the bisector of $\angle BAC$

$I$ is its incentre of $\Delta ABC$.

To prove: $\frac{AI}{ID}=\frac{AB+AC}{BC}$

Proof:

Lets form a diagram, according to question.

Since, in a triangle, the point of concurrence of angular bisectors is called an incentre.

Here $I$ is the incentre of $\Delta ABC$

i.e., In $△ABD,BI$ is is the bisector of $\angle ABD$.

[Vertical angle bisector theorem: "Vertical angle bisector of any angle of a triangle, bisects the opposite side in the same ratio, as that of the other two sides of the triangle.]

By "Vertical angle bisector theorem"

$\frac{AB}{BD}=\frac{AI}{ID}$ ---(1)

In $△ACD$, $CI$ is the bisector of $\angle ACD$.

$\Rightarrow \frac{AC}{CD}=\frac{AI}{ID}$ ---(2)

From (1) & (2)

$\frac{AB}{BD}=\frac{AC}{CD}$

$\Rightarrow \frac{CD}{BD}=\frac{AC}{AB}$

$\Rightarrow \frac{CD}{BD}+1=\frac{AC}{AB}+1$

$\Rightarrow \frac{CD+BD}{BD}=\frac{AC+AB}{AB}$

$\Rightarrow \frac{BC}{BD}=\frac{AC+AB}{AB}$ [Since $DC+BD=BC$]

$\Rightarrow \frac{AB}{BD}=\frac{AC+AB}{BC}$

$\Rightarrow \frac{AI}{ID}=\frac{AC+AB}{BC}$ [Since, from (1) $\frac{AB}{BD}=\frac{AI}{ID}$]

Hence, proved.