Question

The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

Answer 1

Given: In a parallelogram ABCD, diagonals intersect at O and draw a line PQ, which intersects AD and BC.

To prove: PQ divides the parallelogram ABCD into two parts of equal area.

i.e., ar (ABQP) = ar (CDPQ)

Proof: We know that, diagonals of a parallelogram bisect each other.

∴ OA = OC and OB = OD  ...(i)

In ΔAOB and ΔCOD,

OA = OC

OB = OD  ...[From equation (i)]

And ∠AOB = ∠COD   ...[Vertically opposite angles]

∴ ΔAOB = ΔCOD  ...[By SAS congruence rule]

Then, ar (ΔAOB) = ar (ΔCOD)  ...(ii) [Since, congruent figures have equal area]

Now, in ΔAOP and ΔCOQ,

∠PAO = ∠OCQ  ...[Alternate interior angles]

OA = OC  ...[From equation (i)]

And ∠AOP = ∠COQ  ...[Vertically opposite angles]

∴ ΔAOP ≅ ΔCOQ  ...[By ASA congruence rule]

∴ ar (ΔAOP) = ar (ΔCOQ)  ...(iii) [Since, congruent figures have equal area]

Similarly, ar (ΔPOD) = ar (ΔBOQ)  ...(iv)

Now, ar (ABQP) = ar (ΔCOQ) + ar (ΔCOD) + ar (ΔPOD)

= ar (ΔAOP) + ar (ΔAOB) + ar (ΔBOQ) ...[From equations (ii), (iii) and (iv)]

⇒ ar (ABQP) = ar (CDPQ)

Hence proved.