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

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Answer 2

Solution :-

diagonal of a parallelogram divides it into two Triangles of equal area.

∴ ar(ᐃABC) = ar(ᐃACD)

⇒ ar(ABQO) + ar(ᐃCOQ)

= ar(CDPO) + (ᐃAOP) ……[1]

In ᐃAOP and ᐃCOQ, we have ∠AOP = ∠COQ

[vertically opposite angles]

OA = OC [Diagonals of a || gm bisects each other]

∠OAP = ∠OCQ [alternate angles]

∴ ᐃAOP ≅ ᐃCOQ [ASA congruence]

⇒ ar(ᐃAOP) = ar(ᐃCOQ)

[As congruent triangles have equal area] ……[2]

From [1] and [2], ar(ABQO) + ar(ᐃAOP)

= ar(CDPO) + ar(ᐃCOQ)

⇒ ar(ABQP) = ar(CDPQ)

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