Question

$ABCD$ is a parallelogram whose diagonals intersect at $O$. If $P$ is any point on $BO$, prove that:
(i) $ar\left(\Delta ADO\right)=ar\left(\Delta CDO\right)$
(ii) $ar\left(\Delta ABP\right)=ar\left(\Delta CBP\right)$.

Answer 1

$ABCD$ is a parallelogram and $AC,BD$ are diagonals bisect each other.

$\left(i\right)$ Since, Diagonals of a parallelogram bisect each other.

$\therefore$ $AO=OC$

$\Rightarrow$ $O$ is the mid-point of $AC.$

$DO$ is a median of $△DAC$

We know that, a median of a triangles divides it into two triangles of equal areas

$\therefore$ $ar\left(△ADO\right)=ar\left(△CDO\right)$

We know, a median of a triangle divides it into triangles of equal areas.

$\left(ii\right)$ Since, $BO$ is a median of $△BAC,$

$\therefore$ $ar\left(△BOA\right)=ar\left(△BOC\right)$ ---- ( 1 )

$PO$ is a median of $△PAC$

We know, a median of a triangle divides it into triangles of equal areas.

$\therefore$ $ar\left(△POA\right)=ar\left(△POC\right)$ ---- ( 2 )

Subtracting ( 2 ) from ( 1 ) we get,

$\Rightarrow$ $ar\left(△BOA\right)-ar\left(△POA\right)=ar\left(△BOC\right)-ar\left(△POC\right)$

$\Rightarrow$ $ar\left(△ABP\right)=ar\left(△CBP\right)$