Question

in a trapazium ABCD side AB parallel to side CD digonal AC and BD intersect each other at point P. then prove that area of triangle ABP/area of triangle CPD =AB^2/CD^2​

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,

PC

PA

=

PD

PB

PA×PD=PB×PC