Question

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that :

1. ΔAPB is similar to ΔCPD.
2. PA × PD = PB × PC. 

Answer 1

In trapezium ABCD

AB || DC

Diagonals AC and BD intersect each other at P.

To prove:

1. ΔAPB ∼ ΔCPD.
2. PA × PD = PB × PC. 

Proof: In ΔAPB and ΔCPD

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

∠PAB = ∠PCD  ...(Alternate angles)

∴ ΔAPB ∼ ΔCPD  ...(AA axiom)

∴ `(PA)/(PC) = (PB)/(PD)`

`\implies` PA × PD = PB × PC

Hence proved.