In the given figure, ABC is a right angled triangle with ∠BAC = 90°.

(i) Prove that : △ADB ~ △CDA.

(ii) If BD = 18 cm and CD = 8 cm, find AD.

(iii) Find the ratio of the area of △ADB is to area of △CDA.

(i) Let ∠CAD = x.

So, ∠DAB = 90° - x.

In △ABD,

⇒ ∠DAB + ∠ADB + ∠ABD = 180° [By angle sum property]

⇒ 90° - x + 90° + ∠ABD = 180°

⇒ 180° - x + ∠ABD = 180°

⇒ ∠ABD = x + 180° - 180°

⇒ ∠ABD = x.

In △ADB and △CDA,

⇒ ∠CAD = ∠ABD (Both = x)

⇒ ∠CDA = ∠ADB (Both = 90°)

∴ △ADB ~ △CDA [By AA]

Hence, proved that △ADB ~ △CDA.

(ii) Since, △ADB ~ △CDA and corresponding sides of similar triangles are proportional to each other.

$\dfrac{BD}{AD} = \dfrac{AD}{CD}$

⇒ AD² = BD × CD

⇒ AD² = 18 × 8

⇒ AD² = 144

⇒ AD = $\sqrt{144}$ = 12 cm.

Hence, AD = 12 cm.

(iii) We know that,

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

$\therefore\dfrac{\text{Area of ∆ADB}}{\text{Area of ∆CDA}} = \Big(\dfrac{AD}{CD}\Big)^2 \\[1em] = \Big(\dfrac{12}{8}\Big)^2 \\[1em] = \Big(\dfrac{3}{2}\Big)^2 \\[1em] = \dfrac{9}{4}.$

Hence, ratio of the area of △ADB to area of △CDA = 9 : 4.