Question

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

Answer 1

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ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)

${\huge{\mathfrak\pink{prove ;-}}}$

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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Answer 2

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

$\huge\purple{♡}\pink {answer}\purple {♡}$

ABCD is a parallelogram

 diagonals of ∥gm bisect each other

∴ O is m.p of BD i.e., OB = OD

O is m.p of AC i.e; OA = OC

In △ABC;OA=OC

∴BO is the median of △ABC

⇒ar(△AOB)=ar(△BOC)

In △ADC÷ since OA=OC

∴DO is the median of △ADC

⇒ar(△AOD)=ar(△COD)

Similary In △ABD ; OB = OD

∴ AO is the median of △ABD

⇒ar(△AOB)=ar(△AOD)

∴ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)

∴ Proved.

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