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.
Solution
Given: 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(□PDCQ) = ar(□PQBA). Proof: ∵ AC is a diagonal of || gm ABCD ∴ ar(ΔABC) = ar(ΔACD)
To Prove: ar(□PDCQ) = ar(□PQBA). Proof: ∵ AC is a diagonal of || gm ABCD ∴ ar(ΔABC) = ar(ΔACD)
In ΔAOP and ΔCOQ,
AO = CO
| ∵ Diagonals of a parallelogram bisect each other
∠AOP = ∠COQ
| Vertically opposite angles ∠OAP = ∠OCQ
| Alternate interior angles ∴ ΔAOP = ΔCOQ
| By ASA Congruence Rule ∴ ar(ΔAOP) = ar(ΔCOQ)
| ∵ Congruent figures have equal areas ⇒ ar(ΔAOP) + ar(□OPDC)
= ar(ΔCOQ) + ar(□OPDC)
⇒ ar(ΔACD) = ar(□PDCQ)
⇒ 1/2 ar(|| gm ABCD) = ar(□PDCQ)
| From (1) ⇒ ar(□PQBA) = ar(□PDCQ)
⇒ ar(□PDCQ) = ar(□PQBA).
