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(i)

It is given that $\mathrm{AD}$ bisects $\angle \mathrm{A}$.

Applying angle - bisector theorem in $\triangle \mathrm{ABC}$, we get:

$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\Rightarrow \frac{5.6}{\mathrm{DC}}=\frac{6.4}{8}$

$\Rightarrow \mathrm{DC}=\frac{8 \times 5.6}{6.4}=7 \mathrm{~cm}$

(ii)

It is given that $\mathrm{AD}$ bisects $\angle \mathrm{A}$.

Applying angle $-$ bisector theorem in $\triangle \mathrm{ABC}$, we get:

$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

Let $\mathrm{BD}$ be $x \mathrm{~cm}$.

Therefore, $\mathrm{DC}=(6-x) \mathrm{cm}$

$\Rightarrow \frac{x}{6-x}=\frac{10}{14}$

$\Rightarrow 14 x=60-10 x$

$\Rightarrow 24 x=60$

$\Rightarrow x=2.5 \mathrm{~cm}$

Thus, $\mathrm{BD}=2.5 \mathrm{~cm}$

$\mathrm{DC}=6-2.5=3.5 \mathrm{~cm}$

(iii)

It is given that $\mathrm{AD}$ bisects $\angle \mathrm{A}$.

Applying angle $-$ bisector theorem in $\triangle \mathrm{ABC}$, we get:

$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\mathrm{BD}=3.2 \mathrm{~cm}, \mathrm{BC}=6 \mathrm{~cm}$

Therefore, $\mathrm{DC}=6-3.2=2.8 \mathrm{~cm}$

$\Rightarrow \frac{3.2}{2.8}=\frac{5.6}{\mathrm{AC}}$

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

(iv)

It is given that $\mathrm{AD}$ bisects $\angle \mathrm{A}$.

Applying angle $-$ bisector theorem in $\triangle \mathrm{ABC}$, we get :

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

$\Rightarrow \frac{B D}{3}=\frac{5.6}{4}$

$\Rightarrow B D=\frac{5.6 \times 3}{4}$

$\Rightarrow B D=4.2 \mathrm{~cm}$

Hence,  BC = 3 + 4.2 = 7.2 cm