If $ABCD$ is a parallelogram, then prove that $ar(\triangle ABD) = ar(\triangle BCD) = ar(\triangle ABC) = ar(\triangle ACD) = \frac{1}{2}ar( parallelogram\ ABCD)$.

Given:

$ABCD$ is a parallelogram.

To do:

We have to prove that $ar(\triangle ABD) = ar(\triangle BCD) = ar(\triangle ABC) = ar(\triangle ACD) = \frac{1}{2}ar( parallelogram\ ABCD)$.

Solution:

Join $BD$ and $AC$.

We know that,

Diagonals of a parallelogram bisect it into two triangles equal in area.

This implies,

$ar(\triangle ABD) = ar(\triangle BCD) =\frac{1}{2} ar(parallelogram\ ABCD)$...…(i)

$ar(\triangle ABC) = ar(\triangle ADC) = \frac{1}{2} ar(parallelogram\ ABCD)$...…(ii)

From (i) and (ii), we get,

$ar(\triangle ABD) = ar(\triangle BCD) = ar(\triangle ABC) = ar(\triangle ACD) = 

\frac{1}{2} ar(parallelogram\ ABCD)$

Hence proved.

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Updated on: 10-Oct-2022

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