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