In adjoining figure, AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P.

(i) Prove that Δ PAB ∼ Δ PCD.

(ii) Find the length of CD.

(iii) Find the area of Δ PAB : area of Δ PCD.

(i) As we know that,

If two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

∴ PA.PD = PB.PC

$\Rightarrow \dfrac{PA}{PC} = \dfrac{PB}{PD}$

⇒ ∠APB = ∠CPD (Vertically opposite angles)

If the corresponding sides of two triangles are proportional and one angle are equal, then the two triangles are similar.

Hence, proved that Δ PAB ∼ Δ PCD (By SAS rule of similarity).

(ii) Since, Δ PAB ∼ Δ PCD

$\therefore \dfrac{PA}{PC} = \dfrac{AB}{CD}$

Substituting the values, we get :

$\Rightarrow \dfrac{7.5}{5} = \dfrac{9}{CD} \\[1em] \Rightarrow 1.5 = \dfrac{9}{CD} \\[1em] \Rightarrow CD = \dfrac{9}{1.5} \\[1em] \Rightarrow CD = 6.$

Hence, the length of CD = 6 cm.

(iii) As we know that,

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$\therefore \dfrac{\text{Area of ΔPAB}}{\text{Area of ΔPCD}} = \dfrac{PA^2}{PC^2} \\[1em] = \dfrac{(7.5)^2}{5^2} \\[1em] = \dfrac{56.25}{25} \\[1em] = \dfrac{9}{4}$

Hence, the area of ΔPAB : area of ΔPCD = 9 : 4.