In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$. Prove that $ar(AOD) = ar(BOC)$.

(N/A) Given: In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$.
To prove: $ar(AOD) = ar(BOC)$.
Proof:
$1$. Triangles $ADC$ and $BDC$ lie on the same base $CD$ and between the same parallels $AB || CD$.
$2$. Therefore,$ar(ADC) = ar(BDC)$ (Triangles on the same base and between the same parallels are equal in area).
$3$. Subtract $ar(DOC)$ from both sides:
$ar(ADC) - ar(DOC) = ar(BDC) - ar(DOC)$.
$4$. From the figure,$ar(ADC) - ar(DOC) = ar(AOD)$ and $ar(BDC) - ar(DOC) = ar(BOC)$.
$5$. Thus,$ar(AOD) = ar(BOC)$.
Hence proved.