Consider$∆\mathrm{ADE} \mathrm{and} ∆\mathrm{ACB}$

$\angle \mathrm{A} = \angle \mathrm{A}$

Thus, by Angle-Angle similarity, trianles, $∆\mathrm{ACB} ~ ∆\mathrm{ADE}$

(ii) Since, $∆\mathrm{ADE} ~ ∆\mathrm{ACB}$, their sides are proportional.

$\Rightarrow \frac{\mathrm{AE}}{\mathrm{AB}} = \frac{\mathrm{AD}}{\mathrm{AC}} = \frac{\mathrm{DE}}{\mathrm{BC}} ...\left(1\right)$

In $∆\mathrm{ABC}$, by Pythogoras Theorem, we have,

AB² + BC² = AC²

$\Rightarrow {\mathrm{AB}}^2 + 5^2 = {13}^2$

$\Rightarrow \mathrm{AB} = 12\mathrm{cm}$

From equation (1), we get,

$\frac{4}{12} = \frac{\mathrm{AD}}{13} = \frac{\mathrm{DE}}{5}$

$\Rightarrow \frac{1}{3} = \frac{\mathrm{AD}}{12}$

$\text{⇒ AD = }\frac{13}{3} cm$

Also, $\frac{4}{12} = \frac{\mathrm{DE}}{5}$

$\Rightarrow \mathrm{DE} = \frac{20}{12}$

$\Rightarrow \mathrm{DE} = \frac{5}{3} \mathrm{cm}$

(iii) We need to find the area of $∆\mathrm{ADE}$ and quadrilateral BCED.

Area of $∆\mathrm{ADE} = \frac{1}{2} \times \mathrm{AE} \times \mathrm{DE} = \frac{1}{2} \times 4 \times \frac{5}{3} = \frac{10}{3} {\mathrm{cm}}^2$

Area of qudrilateral BCED = Area of $∆\mathrm{ABC}$ - Area of $∆\mathrm{ADE}$

                                     = $\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB} - \frac{10}{3}$

                                     = $\frac{1}{2} \times 5 \times 12 - \frac{10}{3}$

                                     = $30 - \frac{10}{3}$

                                     = $\frac{80}{3} {\mathrm{cm}}^2$

The ratio of areas of $∆\mathrm{ADE}$ to quadrilateral BCED = $\frac{{\displaystyle \frac{10}{3}}}{{\displaystyle \frac{80}{3}}} = \frac{1}{8}$