Question

ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at O and DC at Q. Prove that “ar(POA)=ar(QOC)”

Answer 1

First of all in your question...it should be intersects AB at P..
now we know that a line through O is drawn but it is not parallel to DA and BC.
We have to prove that triangle AOP is congruent to triangle COQ
AO=CO(diagonals of a parallelogram bisect each other)
angleCOQ=angleAOP(v.o.a)
angle OAP=angleQCO(alternate interior angles
by AAS both triangles AOP and COQ are congruent

Answer 2

Here is ur answer mate...

Answer:

it should be intersects AB at P..

now we know that a line through O is drawn but it is not parallel to DA and BC.

We have to prove that triangle AOP is congruent to triangle COQ

AO=CO(diagonals of a parallelogram bisect each other)

angleCOQ=angleAOP(v.o.a)

angle OAP=angleQCO(alternate interior angles

by AAS both triangles AOP and COQ are congruent

   

                      ...Hope it helps u... ^_~