$ABCD$ is a trapezium with $AB || DC$. $A$ line parallel to $AC$ intersects $AB$ at $X$ and $BC$ at $Y$. Prove that $\operatorname{ar}(ADX) = \operatorname{ar}(ACY)$. [Hint: Join $CX$.]

(N/A) We have a trapezium $ABCD$ such that $AB || DC$.
$XY || AC$ meets $AB$ at $X$ and $BC$ at $Y$.
Let us join $CX$.
$\because AB || DC$ [Given]
$\therefore \Delta ADX$ and $\Delta ACX$ are on the same base $AX$ and between the same parallels $AB$ and $DC$.
$\therefore \operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACX) \quad \dots(1)$
$\because AC || XY$
$\therefore \Delta ACX$ and $\Delta ACY$ are on the same base $AC$ and between the same parallels $AC$ and $XY$.
$\therefore \operatorname{ar}(\Delta ACX) = \operatorname{ar}(\Delta ACY) \quad \dots(2)$
From $(1)$ and $(2)$,we have
$\operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACY)$