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

(i) △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) In △ADE and △ACB,

⇒ ∠AED = ∠ABC [Both = 90°]

⇒ ∠EAD = ∠CAB [Common angle]

∴ △ADE ~ △ACB [By AA].

Hence, proved that △ADE ~ △ACB.

(ii) In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ 13² = AB² + 5²

⇒ 169 = AB² + 25

⇒ AB² = 144

⇒ AB = $\sqrt{144}$ = 12 cm.

Since, △ADE ~ △ACB and corresponding sides of similar triangles are proportional to each other.

$\therefore \dfrac{AE}{AB} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{DE}{5} = \dfrac{4}{12} \\[1em] \Rightarrow DE = \dfrac{20}{12} = \dfrac{5}{3} = 1\dfrac{2}{3} \text{ cm}. \\[1em] \Rightarrow \dfrac{AD}{AC} = \dfrac{AE}{AB} \\[1em] \Rightarrow \dfrac{AD}{13} = \dfrac{4}{12} \\[1em] \Rightarrow AD = \dfrac{52}{12} \\[1em] \Rightarrow AD = \dfrac{13}{3} = 4\dfrac{1}{3} \text{ cm}.$

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

(iii) From figure,

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

$= \dfrac{1}{2} \times 4 \times \dfrac{5}{3} \\[1em] = \dfrac{10}{3} \text{ cm}^2 \\[1em] \text{Area of △ABC} = \dfrac{1}{2} \times BC \times AB \\[1em] = \dfrac{1}{2} \times 5 \times 12 \\[1em] = 30 \text{ cm}^2.$

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

$= 30 - \dfrac{10}{3} \\[1em] = \dfrac{90 - 10}{3} \\[1em] = \dfrac{80}{3} \text{ cm}^2. \\[1em] \dfrac{\text{Area of △ADE}}{\text{Area of quadrilateral BCED}} = \dfrac{\dfrac{10}{3}}{\dfrac{80}{3}} \\[1em] = \dfrac{10}{80} = \dfrac{1}{8}.$

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