In the figure,$PQRS$ and $ABRS$ are parallelograms and $X$ is any point on side $BR$. Show that:
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$

(N/A) We have two parallelograms $PQRS$ and $ABRS$ and a point $X$ on $BR$.
$(i)$ To prove that $ar(PQRS) = ar(ABRS)$:
Since parallelogram $PQRS$ and parallelogram $ABRS$ are on the same base $RS$ and between the same parallels $RS$ and $PB$,their areas are equal.
Therefore,$ar(PQRS) = ar(ABRS)$.
$(ii)$ To prove that $ar(AXS) = 1/2 \, ar(PQRS)$:
Since $\Delta AXS$ and parallelogram $ABRS$ are on the same base $AS$ and between the same parallels $AS$ and $BR$,the area of the triangle is half the area of the parallelogram.
Therefore,$ar(AXS) = 1/2 \, ar(ABRS)$.
Since $ar(PQRS) = ar(ABRS)$ (from part $i$),
Substituting this into the equation,we get:
$ar(AXS) = 1/2 \, ar(PQRS)$.