Question

The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5,

Find:

1. ∆APB : ∆CPB
2. ∆DPC : ∆APB
3. ∆ADP : ∆APB
4. ∆APB : ∆ADB

Answer 1

i. Since ΔAPB and ΔCPB have common vertex at B and their bases AP and PC are along the same straight line

`∴(ar(ΔAPB))/(ar(ΔCPB)) = (AP)/(PC) = 3/5`

ii. Since ΔDPC and ΔBPA are similar

∴ `(ar(Δ DPC))/(ar(ΔCPB)) = ((PC)/(AP))^2 = (5/3)^2 = 25/9`

iii. Since ΔADP and ΔAPB have common vertex at A and their bases DP and PB are along the same straight line

∴ `(ar (ΔADP))/(ar(Δ APB)) = (DP)/(PB) = 5/3`

`(ΔAPB ∼ ΔCPD ⇒ (AP)/(PC) = (BP)/(PD) = 3/5)`

iv. Since ΔAPB and ΔADB have common vertex at A and their bases BP and BD are along the same straight line.

∴ `(ar(ΔAPB))/(ar(ΔADB)) = (PB)/(BD) = 3/8`

`(ΔAPB ∼ ΔCPD => (AP)/(PC) = (BP)/(PD) = 3/5 => (BP)/(BD) = 3/8)`