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 : PA x PD = PB x PC. 

Answer 1

In ΔAPB and ΔCPD,

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

∠ABP = ∠CDP ....(alternate angles since AC || DC)

ΔAPB ∼ ΔCPD .....(AA criterion for similarity)

`=> (PA)/(PC) = (PB)/(PD)`  .....(Since corresponding sides of similar triangles are equal)

`=>` PA x PD = PB x PC