In the given figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that △ABC ~ △EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of △BED = 9 cm². Calculate the :

(i) length of AB

(ii) area of △ABC

(i) In △ABC and △EBD,

⇒ ∠EDB = ∠ACB [Given]

⇒ ∠DBE = ∠ABC [Common]

∴ △ABC ~ △EBD

From figure,

⇒ BC = BE + EC = 6 + 4 = 10 cm.

Since, corresponding sides of similar triangles are proportional to each other.

$\Rightarrow \dfrac{AB}{BE} = \dfrac{BC}{BD} \\[1em] \Rightarrow AB = \dfrac{BC \times BE}{BD} \\[1em] \Rightarrow AB = \dfrac{10 \times 6}{5} \\[1em] \Rightarrow AB = 12 \text{ cm}.$

Hence, length of AB = 12 cm.

(ii) We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$\therefore \dfrac{\text{Area of ∆ABC}}{\text{Area of ∆BED}} = \Big(\dfrac{AB}{BE}\Big)^2 \\[1em] \dfrac{\text{Area of ∆ABC}}{9} = \Big(\dfrac{12}{6}\Big)^2 \\[1em] \text{Area of ∆ABC} = \Big(\dfrac{12}{6}\Big)^2 \times 9 \\[1em] \text{Area of ∆ABC} = 36 \text{ cm}^2.$

Hence, area of ∆ABC = 36 cm².