. 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.

i) Let us draw a line segment EF, passing through the point P and parallel to line segment AB in parallelogram ABCD.

AB || EF (By construction)  .....(1)

We know that ABCD is a parallelogram.

∴ AD || BC (Opposite sides of a parallelogram are parallel)

⇒ AE || BF  ..... (2)

From Equations (1) and (2), we obtain

AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

Similarly, it can be deduced that quadrilateral EFCD is a parallelogram.

It can be observed that ΔAPB and parallelogram ABFE is lying on the same base AB and between the same set of parallel lines AB and EF.

∴ Area (ΔAPB) = 1/2 Area (ABFE)  .....(3)

Similarly, for ΔPCD and parallelogram EFCD, Area (ΔPCD) = 1/2 Area (EFCD)  .....(4)

Adding Equations (3) and (4), we obtain Area (ΔAPB) + Area (ΔPCD) = 1/2 [Area (ABFE) + Area (EFCD)]

Area (ΔAPB) + Area (ΔPCD) = 1/2 Area (ABCD)  .....(5)