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,

∠BAP = ∠DCP   ...(∠s on same segment)

∠ABP = ∠CDP   ...(∠s on same segment)

∴ ΔAPB ~ ΔCPD   ...(AA axiom)

b. ${\displaystyle \frac{AB}{CD}=\frac{3}{2}}$

∴ CD = 6 cm

c. ${\displaystyle \frac{Area(\Delta APB)}{Area \Delta CPD}=\frac{B{P}^2}{D{P}^2}}$

= ${\displaystyle \frac{9}{4}}$

${\displaystyle \Rightarrow }$ 9 : 4