Question

P is a point in the interior of a parallelogram ABCD show that ar (triangle APB)+ar (triangle PCB)=1/2 ar (ABCD)

Answer 1

(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD,

AB || EF (By construction) ... (1)

ABCD is a parallelogram.

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

⇒ AE || BF ... (2)

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

AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

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

∴ Area (ΔAPB) =
$\frac{1}{2}$

Area (ABFE) ... (3)

Similarly, for ΔPCD and parallelogram EFCD,

Area (ΔPCD) =

$\frac{1}{2}$

Area (EFCD) ... (4)

Adding equations (3) and (4), we obtain