Question

In the given figure (drawn not to scale) chords AD and BC intersect at P, where AB = 9 cm, PB = 3 cm and PD = 2 cm.

1. Prove that ΔAPB ~ ΔCPD.
2. Find the length of CD. 
3. Find area ΔAPB : area ΔCPD.

Answer 1

a. In ΔAPB and ΔCPD,

∠APB = ∠CPD   ...(Vertically opposite angles are equal.)

∠BAP = ∠DCP   ...(Angles in same segment are equal.)

∴ ΔAPB ~ ΔCPD   ...(By AA axiom)

b. We know that corresponding sides of similar triangles are proportional.

∴ `(CD)/(AB) = (PD)/(PB)`

`(CD)/9 = 2/3`

CD = `9 xx 2/3`

CD = 6 cm

c. We know that the ratio of the area of similar triangles is equal to the ratio of the square of the corresponding sides.

∴ `("Area of ΔAPB")/("Area of ΔCPD") = (PB^2)/(PD^2)`

= `3^2/2^2`

= `9/4`

`\implies` 9 : 4