Question

In the given Figure, diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$ such that $OB=OD$. If $AB=CD$, then show that :
(i) $ar\left(DOC\right)=ar\left(AOB\right)$
(ii) $ar\left(DCB\right)=ar\left(ACB\right)$
(iii) $DA\parallel CB$ or $ABCD$ is a parallelogram.

Answer 1

Construction:

Draw $DE\perp AC$ and $BF\perp AC$.

In $△DEO$ and $△BFO$

$\angle DEO=\angle BFO$ ....By construction

$\angle DOE=\angle BOF$ ....Vertically opposite angles
$OD=OB$ .....Given

$△DOE\cong △BOF$ ...SAA test of congruence ....(1)

$\therefore DE=BF$ .... CSCT ....(2)

and

$OE=OF$ ... CSCT ....(3)

In $△DCE$ and $△BFA$

$\angle DEC=\angle BFA$ ..... By construction

$DC=AB$ ....Given
$DE=BF$ ....From (2)

$△DEC\cong △BFA$ .....(RHS) Right Hypotenuse Side test ....(4)

$\therefore EC=AF$ ......C.S.C.T. ....(5)

Similarly, we can prove $△DOA\cong △BOC$ ....(6)

Adding (1) and (2), we get

$OE+EC=OF+AF$

$\therefore OC=OA$ .....(7)

(i) Adding (1) & (4), we get,

$A\left(△DOE\right)+A\left(△DEC\right)=A\left(△BOF\right)+A\left(△BFA\right)$

$\therefore A\left(△DOC\right)=A\left(△BOA\right)$ ....(8)

(ii) Adding (6) and (8), we get

$A\left(△DOC\right)+A\left(△DOA\right)=A\left(△BOA\right)+A\left(△BOC\right)$
$\Rightarrow A\left(△ACD\right)=A\left(△ACB\right)$

(iii) So, from (7) and given,

$\square ABCD$ is a parallelogram .....(Diagonals bisect each other)

and $DA\parallel CB$ ....Opp. sides of parallelogram are parallel to each other