CBSE Class 9 Answered
Diagonals of a parallelogram ABCD intersect
at 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.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
Expert Answer
Since diagonals of parallelogram divide it into two triangles of equal area.
Area(
ABC) = area(ACD)
area (quad ABQO) + (COQ) = area(quad. CDPO) + area(AOP)
...(1)
In
AOP and COQ,
AOP
= COQ (vert. opp. Angles)
OA = OC(diagonals of a||gm bisect each other)
OAP
= OCQ (alternate angles)
(ASA congruence)
....(2)
From (1) and (2), we have:
area(quad.
ABQO) + area(
) = area(quad. CDPO) + area(
)
Hence, area(quad. ABQP) = area(quad. CDQP)
Answered by | 04 Jun, 2014, 03:23: PM
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