Given: In parallelogram ABCD, P & Q any twopoints lying on the sides DC and AD.

To show:

ar (APB) = ar (BQC).Proof:

Here, ΔAPB and ||gm ABCD stands on the samebase AB and lie between same parallel AB and DC.

Therefore,

ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)

Similarly,

Parallelogram ABCD and ∆BQC stand on the samebase BC and lie between the same parallel BC and AD.

ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)

From eq (i) and (ii),

we have

ar(ΔAPB)= ar(ΔBQC)

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Given: ABCD is a parallelogram

So, AB||CD & AD|| BC

To show:

(i) ar (APB) + ar (PCD) = ar (ABCD)

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

Proof:

(i)

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 thesame 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 thesame 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)

(ii)

A line EF is drawn parallelto AD passing through P.

In a parallelogram,

AD || EF (by construction) — (i)

Thus,

AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on thesame base AD and between the same parallel lines AD and EF.

∴ar(ΔAPD) = 1/2 ar(AEFD) — (iii)

also,

In ΔPBC and parallelogram BCFE are lying on thesame base BC and between the same parallel lines BC and EF.

∴ar(ΔPBC) = 1/2 ar(BCFE) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPD)+ ar(ΔPBC) = 1/2 {ar(AEFD)+ar(BCFE)}

ar(ΔAPD)+ ar(ΔPBC) =1/2ar(ABCD)

ar(APD) +ar(PBC) = ar(APB) + ar(PCD) ( From parti)

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