Question

DIAGONALS OF PARALLELOGRAM abcd INTERSECT AT A POINT I A LINE IS DRAWN TO INTERSECT ad AT p AND bc AT o SHOW THAT pq DIVIDES THE PARALLELOGRAM INTO TWO EQUAL AREAS

Answer 1

Done some changes intersecting point is taken as o

Answer 2

Step-by-step explanation:

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

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