Show that the diagonals of a parallelogram divide it into four triangles of equal area.
To do:
We have to show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
Let in a parallelogram $ABCD$ diagonals $AC$ and $BD$ intersect at $O$.
We know that,
The diagonals of a parallelogram bisect each other.
This implies,
$OA = OC$ and $OB = OD$.
The median of a triangle divides it into two triangles of equal areas.
In $\triangle ABC$, $BO$ is the median.
This implies,
$ar (\triangle OAB) = ar (\triangle OBC)$.........(i)
In $\triangle ABD$, $AO$ is the median.
This implies,
$ar (\triangle OAB) = ar (\triangle OAD)$.......(ii)
In $\triangle ACD$, $DO$ is the median.
This implies,
$ar (\triangle OAD) = ar (\triangle OCD)$.........(iii)
From (i), (ii) and (iii), we get,
$ar (\triangle OAB) = ar (\triangle OBC) = ar (\triangle OCD) = ar(\triangle OAD)$
Hence proved.
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