In the figure,diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$ such that $OB = OD$. If $AB = CD$,then show that:
$(i)$ $ar(DOC) = ar(AOB)$
$(ii)$ $ar(DCB) = ar(ACB)$
$(iii)$ $DA \parallel CB$ or $ABCD$ is a parallelogram.
[Hint: From $D$ and $B$,draw perpendiculars to $AC$.]

(A) We have a quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at $O$.
We are given that $OB = OD$ and $AB = CD$.
Let us draw $DE \perp AC$ and $BF \perp AC$.
$(i)$ To prove that $ar(\Delta DOC) = ar(\Delta AOB)$:
In $\Delta DEO$ and $\Delta BFO$,we have:
$DO = BO$ (Given)
$\angle DOE = \angle BOF$ (Vertically opposite angles)
$\angle DEO = \angle BFO = 90^\circ$ (By construction)
Therefore,$\Delta DEO \cong \Delta BFO$ ($AAS$ congruence rule).
This implies $DE = BF$ and $ar(\Delta DEO) = ar(\Delta BFO)$ $(1)$.
Now,in $\Delta DEC$ and $\Delta BFA$,we have:
$\angle DEC = \angle BFA = 90^\circ$
$DE = BF$ (From above)
$DC = BA$ (Given)
Therefore,$\Delta DEC \cong \Delta BFA$ ($RHS$ congruence rule).
This implies $ar(\Delta DEC) = ar(\Delta BFA)$ $(2)$.
Adding $(1)$ and $(2)$,we get:
$ar(\Delta DEO) + ar(\Delta DEC) = ar(\Delta BFO) + ar(\Delta BFA)$
$ar(\Delta DOC) = ar(\Delta AOB)$.
$(ii)$ To prove that $ar(DCB) = ar(ACB)$:
Since $ar(\Delta DOC) = ar(\Delta AOB)$,
Adding $ar(\Delta BOC)$ on both sides,we have:
$ar(\Delta DOC) + ar(\Delta BOC) = ar(\Delta AOB) + ar(\Delta BOC)$
$ar(\Delta DCB) = ar(\Delta ACB)$.
$(iii)$ To prove that $DA \parallel CB$:
Since $\Delta DCB$ and $\Delta ACB$ are on the same base $CB$ and have equal areas,they must lie between the same parallels.
Therefore,$DA \parallel CB$. Since $DA \parallel CB$ and $AB = CD$ (with $OB=OD$),$ABCD$ is a parallelogram.