show that the diagonals of a parralalogram divide it into four triangles of equal area.
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To prove:- ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)
Proof:-
Let ABCD be a parallelogram with diagonals AC and BD
intersecting at O. Since the diagonals of a parallelogram bisect each other at the point of intersection.
Therefore,
AO=OC and BO=OD
We know that the median of a triangle divides it into two equal parts.
Now,
In △ABC,
∵BO is median.
ar(△AOB)=ar(△BOC).....(1)
In △BCD,
∵CO is median.
ar(△BOC)=ar(△COD).....(2)
In △ACD,
∵DO is median.
ar(△AOD)=ar(△COD).....(3)
From equation (1),(2)&(3), we get
ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)
Hence proved.
solution
Step-by-step explanation:
hope it's helpful for you....
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