In parallelogram $ABCD$,diagonals $AC$ and $BD$ intersect at point $O$. Point $P$ lies on line segment $BO$. Prove that,$ar(ABP) = ar(CBP)$.

(A) $1$. In a parallelogram,the diagonals bisect each other. Therefore,$O$ is the midpoint of $BD$,which implies $BO = OD$.
$2$. Consider $\triangle ABD$ and $\triangle CBD$. Since $ABCD$ is a parallelogram,$AB = CD$ and $AD = BC$. Also,$BD$ is a common diagonal.
$3$. In $\triangle ABD$ and $\triangle CBD$,the median from vertex $A$ to $BD$ is $AO$. Since $O$ is the midpoint of $BD$,$AO$ divides $\triangle ABD$ into two triangles of equal area: $ar(ABO) = ar(ADO)$.
$4$. Similarly,in $\triangle CBD$,$CO$ is the median to $BD$,so $ar(CBO) = ar(CDO)$.
$5$. However,the problem asks to prove $ar(ABP) = ar(CBP)$. Let us consider the triangles $\triangle ABP$ and $\triangle CBP$. These triangles share the same base $BP$ on the line $BD$.
$6$. The height of $\triangle ABP$ with respect to base $BP$ is the perpendicular distance from $A$ to $BD$,let it be $h_1$. The height of $\triangle CBP$ with respect to base $BP$ is the perpendicular distance from $C$ to $BD$,let it be $h_2$.
$7$. In a parallelogram,the distance from opposite vertices to a diagonal is equal. Thus,$h_1 = h_2$.
$8$. Since $ar(ABP) = \frac{1}{2} \times BP \times h_1$ and $ar(CBP) = \frac{1}{2} \times BP \times h_2$,and $h_1 = h_2$,it follows that $ar(ABP) = ar(CBP)$.