Question

$\huge\bold\blue{QuEtIoN}$

Show that the diagonals of a parallelogram divide it into four triangles of equal area.

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$\huge\boxed{\fcolorbox{black}{yellow}{DON'T\:SPAM\: ALLOWED}}$​

Answer 1

$\huge\bold\blue{Solution}$

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.

Answer 2

$\huge\rm{\underline{\red{Answer}}}$

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.

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