Question

in the adjoining figure, P is a point in the interior of the parallelogram ABCD. show that ar(∆ APB) + ar(∆ PCD) =1/2 ar ( || GM ABCD ) .

Answer 1

Hey mate here is your answer

Given: ABCD is a parallelogram

So, AB||CD & AD|| BC

To show:

(i) ar (APB) + ar (PCD) = ar (ABCD) 

Proof:

Through the point P ,draw GH parallel to AB.

In a parallelogram,

AB || GH (by construction) — (i)

Thus,

AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

In ΔAPB and parallelogram ABHG are lying on the

same base AB and between the same parallel lines AB and GH.

∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)

also,

In ΔPCD and parallelogram CDGH are lying on the

same base CD and between the same parallel lines CD and GH.

∴ ar(ΔPCD) = 1/2 ar(CDGH) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) +

ar(CDGH)}

 ar(APB) +

ar(PCD) = 1/2 ar(ABCD)

HOPE IT HELPED YOU