(i) A line GH is drawn parallel to AB passing through P.

In a parallelogram,

AB || GH (by construction) — (i)

∴ AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

ΔAPB and parallelogram ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.

∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)

ΔPCD and parallelogram CDGH are lying on the same base CD and in-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 parallel to AD passing through P.

In the parallelogram,

AD || EF (by construction) — (i)

∴ AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

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

∴ar(ΔAPD) = 1/2 ar(AEFD) — (iii)

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

∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)

Adding equations (iii) and (iv)

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

⇒ ar(APD)+ar(PBC) = ar(APB) + ar(PCD)