Total Questions :36

In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]

Answer

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

(ii) A line EF is drawn parallel to 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 the same 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 the same 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(APD) + ar(PBC) = ar(APB) + ar(PCD)