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Question -

Inthe given figure, P is a point in the interior of a parallelogram ABCD. Showthat

(i)ar (APB) + ar (PCD) =  ar (ABCD)

(ii)ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint:Through. P, draw a line parallel to AB]



Answer -

Let us draw a line segmentMN, passing through point P and parallel to line segment AD.

In parallelogram ABCD,

MN || AD (By construction)… (6)

ABCD is a parallelogram.

∴AB || DC (Opposite sides of a parallelogram)

⇒AM || DN … (7)

From equations (6) and(7), we obtain

MN || AD and AM || DN

Therefore, quadrilateralAMND is a parallelogram.

It can be observed thatΔAPD and parallelogram AMND are lying on the same base AD and between the sameparallel lines AD and MN.

∴ Area (ΔAPD) = Area (AMND) … (8)

Similarly, for ΔPCB andparallelogram MNCB,

Area(ΔPCB) = Area (MNCB) … (9)
Addingequations (8) and (9), we obtain

On comparing equations (5)and (10), we obtain

Area (ΔAPD) + Area (ΔPBC)= Area (ΔAPB) + Area (ΔPCD)

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