ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC.

(i) Prove that △ADE ~ △ACB.

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

(iii) Find, area of △ADE : area of quadrilateral BCED.

(i) Considering △ADE and △ACB.

∠A = ∠A (Common angles)

∠AED = ∠ABC (Both are equal to 90°)

Hence, by AA axiom △ADE ~ △ACB.

(ii) △ABC is a right angled triangle.

By pythagoras theorem,

$AC^2 = AB^2 + BC^2 \\[1em] AB^2 = AC^2 - BC^2 \\[1em] AB^2 = 13^2 - 5^2 \\[1em] AB^2 = 169 - 25 \\[1em] AB^2 = 144 \\[1em] AB = \sqrt{144} \\[1em] AB = 12 \text{ cm}.$

Since triangles are similar hence the ratio of their corresponding sides are equal.

$\therefore \dfrac{AE}{AB} = \dfrac{AD}{AC} \\[1em] \Rightarrow \dfrac{4}{12} = \dfrac{AD}{13} \\[1em] \Rightarrow AD = \dfrac{4 \times 13}{12} \\[1em] \Rightarrow AD = \dfrac{13}{3} \\[1em] \Rightarrow AD = 4\dfrac{1}{3}.$

Similarly,

$\therefore \dfrac{AE}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{4}{12} = \dfrac{DE}{5} \\[1em] \Rightarrow DE = \dfrac{4 \times 5}{12} \\[1em] \Rightarrow DE = \dfrac{5}{3} \\[1em] \Rightarrow DE = 1\dfrac{2}{3}.$

Hence, the length of AD = $4\dfrac{1}{3}$ cm and of DE = $1\dfrac{2}{3}$ cm.

(iii) Area of a right angled triangle is given by

$\dfrac{1}{2} \times \text{Base} \times \text{Height}$.

Area of △ADE = $\dfrac{1}{2} \times AE \times DE$

$= \dfrac{1}{2} \times 4 \times \dfrac{5}{3} \\[1em] = \dfrac{10}{3} \text{ cm}^2.$

Area of quadrilateral BCED = Area of △ABC - Area of △ADE

$= \dfrac{1}{2} \times BC \times AB - \dfrac{10}{3} \\[1em] = \dfrac{1}{2} \times 5 \times 12 - \dfrac{10}{3} \\[1em] = 30 - \dfrac{10}{3} \\[1em] = \dfrac{90 - 10}{3} \\[1em] = \dfrac{80}{3} \text{ cm}^2$

Hence, area of △ADE : area of quadrilateral BCED is

$= \dfrac{\dfrac{10}{3}}{\dfrac{80}{3}} \\[1em] = \dfrac{10 \times 3}{80 \times 3} \\[1em] = \dfrac{1}{8} \\[1em] = 1 : 8.$

Hence, area of △ADE : area of quadrilateral BCED is 1 : 8.