Question

is a point in the interior of a parallelogram ABCD show that
(i) ar (APB)+ar(PCD)=1/2ar(ABCD)

Answer 1

(i) Let us draw a line segment EF, passing through point P andparallel to line segment AB.

Inparallelogram ABCD,

AB|| EF (By construction) ... (1)

ABCDis a parallelogram.

∴AD || BC (Opposite sides of a parallelogram)

⇒AE || BF ... (2)

Fromequations (1) and (2), we obtain

AB|| EF and AE || BF

Therefore,quadrilateral ABFE is a parallelogram.

Itcan be observed that ΔAPBand parallelogram ABFE are lying on the same base AB and between thesame parallel lines AB and EF.

∴Area (ΔAPB) = Area(ABFE) ... (3)

Similarly,for ΔPCD and parallelogramEFCD,

Area(ΔPCD) = Area(EFCD) ... (4)

Addingequations (3) and (4), we obtain

i) ar (APB)+ar(PCD)=1/2ar(ABCD)