Question

In trapezium ABCD, AB parallel to CD and its diagonals intersect at P. If AB= 6cm, and DC=3cm, then the ratio of the areas of triangle APB and triangle CPD is

Answer 1

Answer:

In △ APB and △ CPD,

∠APB=∠CPD (Vertically opposite angles)

∠ABP=∠CDP (Alternate angles of parallel sides AB and CD)

∠BAP=∠DCP (Alternate angles of parallel sides AB and CD)

Hence, △APB∼△CPD (AAA rule)

Thus, PCPA=PDPB

PA×PD=PB×PC