If a ∆ ABC, AD is the bisector of ∠A, meeting side BC at D.

Solution:

(i) It is given that $B D=2.5 \mathrm{~cm}, A B=5 \mathrm{~cm}$ and $A C=4.2 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.

We have to find $D C$.

Since $A D$ is $\angle A$ bisector

Then $\frac{A B}{A C}=\frac{2.5}{D C}$

$\frac{5}{4.2}=\frac{2.5}{D C}$

$5 D C=4.2 \times 2.5$

$D C=\frac{4.2 \times 2.5}{5}$

$=2.1$

Hence $D C=2.1 \mathrm{~cm}$

(ii) It is given that $B D=2 \mathrm{~cm}, A B=5 \mathrm{~cm}$ and $D C=3 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.

We have to find $A C$.

Since $A D$ is $\angle A$ bisector

So $\frac{A B}{A C}=\frac{B D}{D C}(A D$ is bisector of $\angle A$ and side $B C)$

Then

$\frac{5}{A C}=\frac{2}{3}$

$\Rightarrow 2 A C=5 \times 3$

$\Rightarrow A C=\frac{15}{2}$

$=7.5$

Hence $A C=7.5 \mathrm{~cm}$

(iii) It is given that $A B=3.5 \mathrm{~cm}, A C=4.2 \mathrm{~cm}$ and $D C=2.8 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$

We have to find $B D$.

Since $A D$ is $\angle A$ bisector

So $\frac{A B}{A C}=\frac{B D}{D C}(A D$ is bisector of $\angle A$ and side $B C)$

Then

$\frac{3.5}{4.2}=\frac{B D}{2.8}$

$\Rightarrow B D=\frac{3.5 \times 2.8}{42}$

$\Rightarrow B D=\frac{7}{3}$

$=2.3$

Hence $B D=2.3 \mathrm{~cm}$

(iv) It is given that $A B=10 \mathrm{~cm}, A C=14 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.

We have to find $B D$ and $D C$.

Since $A D$ is $\angle A$ bisector

So $\frac{A B}{A C}=\frac{B D}{D C}\left(A D^{\text {is bisector of }} \angle A\right.$ and side $\left.B C\right)$

Let $\mathrm{BD}=x \mathrm{~cm}$. Then $\mathrm{CD}=(6-x) \mathrm{cm}$

Then,

1014=x6-x⇒14x=60-10x⇒24x=60⇒x=6024=2.5

Hence, BD = 2.5 cm and DC = 6 − 2.5 = 3.5 cm.

(v) It is given that $A C=4.2 \mathrm{~cm}, D C=6 \mathrm{~cm}$ and $B C=10 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.

We have to find $A B$.

Since $A D$ is $\angle A$ bisector

$S O \frac{A C}{A B}=\frac{D C}{B D}$

Then

$\frac{4.2}{A B}=\frac{6}{4}$

$\Rightarrow 6 A B=4.2 \times 4$

$\Rightarrow A B=\frac{4.2 \times 4}{6}$

$=\frac{16.8}{6}$

Hence $A B=2.8 \mathrm{~cm}$

(vi) It is given that $A B=5.6 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $D C=3 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.

We have to find $B C$.

Since $A D$ is $\angle A$ bisector

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

Then

$\frac{6}{5.6}=\frac{3}{D C}$

$\Rightarrow D C=2.8$

So

$B C=2.8+3$

$=5.8$

Hence $B C=5.8 \mathrm{~cm}$

(vii) If it is given that $A B=5.6 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $B D=3.2 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$

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

$\frac{5.6 \mathrm{~cm}}{A C}=\frac{3.2 \mathrm{~cm}}{2.8 \mathrm{~cm}} \quad[D C=B C-B D]$

$A C=\frac{5.6 \times 2.8}{3.2} \mathrm{~cm}=4.9 \mathrm{~cm}$

(viii) It is given that $A B=10 \mathrm{~cm}, A C=6 \mathrm{~cm}$ and $B C=12 \mathrm{~cm}$.

In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.

We have to find $B D$ and $D C$.

Since $A D$ is $\angle A$ bisector

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

Let BD = x cm

Then

$\frac{6}{10}=\frac{12-x}{x}$

$\Rightarrow 6 x=120-10 x$

$\Rightarrow 16 x=120$

$\Rightarrow x=\frac{120}{16}$

$\Rightarrow x=7.5$

Now 

$D C=12-B D$

$=12-7.5$

$=4.5$

Hence $B D=7.5 \mathrm{~cm}$ and $D C=4.5 \mathrm{~cm}$