In the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.

(N/A) Given: $CD \parallel AE$ and $CY \parallel BA$.
To prove: $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
Proof:
Since $\triangle ABC$ and $\triangle ABY$ are on the same base $AB$ and between the same parallels $CY \parallel BA$,their areas are equal:
$\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle ABY)$
Subtracting $\operatorname{ar}(\triangle ABX)$ from both sides:
$\operatorname{ar}(\triangle ABC) - \operatorname{ar}(\triangle ABX) = \operatorname{ar}(\triangle ABY) - \operatorname{ar}(\triangle ABX)$
From the figure,$\operatorname{ar}(\triangle ABC) - \operatorname{ar}(\triangle ABX) = \operatorname{ar}(\triangle CBX)$ and $\operatorname{ar}(\triangle ABY) - \operatorname{ar}(\triangle ABX) = \operatorname{ar}(\triangle AXY)$.
Therefore,$\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
Hence proved.