Given, ABCD and AEFD are two parallelograms.
To prove `" " ar(DeltaPEA) = ar (DeltaQFD)`
Proof In quadrilateral PQDA,
`AP|| DQ" "` [since, in parallelogram ABCD, `AB ||CD`]
and `" " PQ || AD" "` [since, in parallelogram AEFD, `FE || AD`]
Then, quadrilateral PQDA is a parallelogram.
Also, parallelogram PQDA and AEFD are on the same base AD and between the same parallels AD and EQ.
`therefore" "` ar (parallelogram PQDA) = ar (parallelogram AEFD)
On subtracting ar (quadrilateral APFD) from both sides, we get
ar (parallelogram PQDA)- ar (quadrilateral APFD)
= ar (parallelogram AEFD) - ar (quadrilateral APFD)
`rArr" "` `ar (DeltaQFD) = ar (DeltaPEA)" "` Hence proved.
