In Fig. 4.60, check whether AD is the bisector of ∠A of ∆ABC in each of the following:

Solution:

(i) It is given that $A B=5 \mathrm{~cm}, A C=10 \mathrm{~cm}, B D=1.5 \mathrm{~cm}$ and $C D=3.5 \mathrm{~cm}$.

We have to check whether $A D$ is bisector of $\angle A$.

First we will check proportional ratio between sides.

Now

ABAC=510=12

BDCD=1.53.5=37

Since ABAC\neqBDCD

Hence $A D$ is not the bisector of $\angle A$.

(ii) It is given that, $A B=4 \mathrm{~cm}, A C=6 \mathrm{~cm}, B D=1.6 \mathrm{~cm}$ and $C D=2.4 \mathrm{~cm}$.

We have to check whether $A D$ is bisector of $\angle A$.

First we will check proportional ratio between sides.

So $\frac{A B}{A C}=\frac{B D}{D C}$

$\Rightarrow 46=1.62 .4 \Rightarrow 23=23$

(It is proportional)

Hence, $A D$ is bisector of $\angle A$.

(iii) It is given that, $A B=8 \mathrm{~cm}, A C=24 \mathrm{~cm}, B D=6 \mathrm{~cm}$ and $B C=24 \mathrm{~cm}$.

We have to check whether $A D$ is bisector of $\angle A$.

First we will check proportional ratio between sides.

Now

$D C=B C-B D$

$D C=24-6$

$=18$

So $\frac{A B}{A C}=\frac{B D}{D C}$

$\Rightarrow \frac{8}{24}=\frac{6}{18}$

$\Rightarrow \frac{1}{3}=\frac{1}{3}$

(It is proportional)

Hence, $A D$ is bisector of $\angle A$.

(iv) It is given that, AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm.

We have to check whether $A D$ is bisector of $\angle A$.

First we will check proportional ratio between sides.

So $\frac{A B}{A C}=\frac{B D}{D C}$

⇒68=1.52⇒34=34

(It is proportional)

Hence $A D$ is bisector of $\angle A$.

(v) It is given that AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

We have to check whether $A D$ is bisector of $\angle A$.

First we will check proportional ratio between sides.

Now

ABAC=512

BDCD=2.59=518

Since ABAC\neqBDCD

Hence $A D$ is not the bisector of $\angle A$.