Question

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Answer 1

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) = ½ ar(ABHG) — (iii)

also,

ΔPCD and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.

∴ ar(ΔPCD) = ½ ar(CDGH) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = ½ [ar(ABHG)+ar(CDGH)]

⇒ ar(APB)+ ar(PCD) = ½ 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.

Now,

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

∴ar(ΔAPD) = ½ ar(AEFD) — (iii)

also,

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

∴ar(ΔPBC) = ½ ar(BCFE) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPD)+ ar(ΔPBC) = ½ {ar(AEFD)+ar(BCFE)}

⇒ar(APD)+ar(PBC) = ar(APB)+ar(PCD)

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