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

According to the 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.

To Prove:

Ar (parallelogram PDCQ) = ar (parallelogram PQBA).

Proof:

AC is a diagonal of || gm ABCD

∴ ar(ΔABC) = ar(ΔACD)

= ½ ar (||gm ABCD) …(1)

In ΔAOP and ΔCOQ,

AO = CO

Since, diagonals of a parallelogram bisect each other,

We get,

∠AOP = ∠COQ

∠OAP = ∠OCQ (Vertically opposite angles)

∴ ΔAOP = ΔCOQ (Alternate interior angles)

∴ ar(ΔAOP) = ar(ΔCOQ) (By ASA Congruence Rule)

We know that,

Congruent figures have equal areas

So,

ar(ΔAOP) + ar(parallelogram OPDC) = ar(ΔCOQ) + ar(parallelogram OPDC)

⇒ ar(ΔACD) = ar(parallelogram PDCQ)

⇒ ½ ar(|| gm ABCD) = ar(parallelogram PDCQ)

From equation (1),

We get,

ar(parallelogram PQBA) = ar(parallelogram PDCQ)

⇒ ar(parallelogram PDCQ) = ar(parallelogram PQBA).

Hence Proved.