In the given figure,$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on extended $BC$. Prove that $ar(BDE) = ar(ACED)$.

(N/A) Given: $ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on $BC$ produced.
To prove: $ar(BDE) = ar(ACED)$.
Proof:
Triangles $ADC$ and $BDC$ lie on the same base $DC$ and between the same parallels $AB$ and $DC$.
Therefore,$ar(ADC) = ar(BDC)$.
Adding $ar(DCE)$ to both sides of the equation:
$ar(ADC) + ar(DCE) = ar(BDC) + ar(DCE)$
From the figure,$ar(ADC) + ar(DCE) = ar(ACED)$ and $ar(BDC) + ar(DCE) = ar(BDE)$.
Therefore,$ar(ACED) = ar(BDE)$ or $ar(BDE) = ar(ACED)$.
Hence proved.