In parallelogram $ABCD$,diagonals $AC$ and $BD$ intersect at point $O$. Point $P$ lies on line segment $BO$. Prove that,$ar(ADO) = ar(CDO)$.
(N/A) $1$. In a parallelogram,the diagonals bisect each other. Therefore,$O$ is the midpoint of $AC$.
$2$. Consider $\triangle ADC$. Since $O$ is the midpoint of $AC$,$DO$ is the median of $\triangle ADC$.
$3$. $A$ median of a triangle divides it into two triangles of equal area.
$4$. Therefore,$ar(ADO) = ar(CDO)$.
$2$. Consider $\triangle ADC$. Since $O$ is the midpoint of $AC$,$DO$ is the median of $\triangle ADC$.
$3$. $A$ median of a triangle divides it into two triangles of equal area.
$4$. Therefore,$ar(ADO) = ar(CDO)$.
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