In the figure,$BD \parallel CA$,$E$ is the mid-point of $CA$,and $BD = \frac{1}{2} CA$. Prove that $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.

(N/A) Given: $BD \parallel CA$,$E$ is the mid-point of $CA$,so $CE = \frac{1}{2} CA$.
Since $BD = \frac{1}{2} CA$,we have $BD = CE$.
Also,$BD \parallel CE$ (as $BD \parallel CA$).
Since one pair of opposite sides is equal and parallel,$BCED$ is a parallelogram.
Now,$\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$ because they lie on the same base $BC$ and between the same parallels $BC$ and $DE$.
In $\triangle ABC$,$BE$ is a median because $E$ is the mid-point of $CA$.
$A$ median of a triangle divides it into two triangles of equal area.
Therefore,$\operatorname{ar}(EBC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(EBC)$.
Substituting $\operatorname{ar}(EBC) = \operatorname{ar}(DBC)$,we get $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.
Hence proved.