Question

In fig. 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|| CB or ABCD is parallelogram.

Answer 1

$\left(i\right)$Let us draw $DP$perpendicular to $AC$ and $BQ$ perpendicular $AC$

In $△DOP$ and $△BOQ$

$\angle DPO=\angle BQO={90}^{\circ }$

$\angle DOP=\angle BOQ$ (vertically opposite angles)

$OD=OB$(given)

$△DOP\cong △BOQ$(AAS congruence rule)

$\therefore DP=BQ$ by CPCT .....$\left(1\right)$

and area$\left(DOP\right)=$area$\left(BOQ\right)$ ......$\left(2\right)$ since area of congruent triangles are equal.

In $△CDP$ and $△ABQ$,

$\angle CPD=\angle AQB={90}^{\circ }$

$CD=AB$(given)

$DP=BQ$ from $\left(1\right)$

$\therefore △CDP\cong △ABQ$ by RHS congruence rule.

$\Rightarrow$ area$△CPD=$area$△AQB$ ........$\left(3\right)$since area of congruent triangles is equal.

Adding $\left(2\right)$ and $\left(3\right)$

Area$△\left(DOP\right)+$Area$△\left(CDP\right)=$area$△\left(BOQ\right)+$area$△\left(ABQ\right)$

$\Rightarrow$ Area$△\left(DOC\right)=$Area$△\left(AOB\right)$

Hence proved.

In part $\left(i\right)$ we proved that

Area$△\left(DOC\right)=$Area$△\left(AOB\right)$

Adding Area$△\left(OCB\right)$ both sides,

Area$△\left(DOC\right)+$Area$△\left(OCB\right)=$Area$△\left(AOB\right)+$Area$△\left(OCB\right)$

$\Rightarrow$ Area$△\left(DCB\right)=$area$△\left(ACB\right)$

In part $\left(ii\right)$ we proved

Area$△\left(DCB\right)=$area$△\left(ACB\right)$

We know that two triangles having same base and equal areas, lie between same parallels.

Here $△DBC$ and $△ABC$ are on the same base $BC$ and are equal in area,

So, these triangles lie between the same parallels $DA$ and $CB$

$\therefore DA\parallel CB$

Now, in $△DOA$ and $△BOC$

$\angle DOA=\angle BOC$(vertically opposite angles)

$\angle DAO=\angle BOC$ since $DA\parallel CB,AC$ transversal, alternate angles are equal.

$OD=OB$(given)

$\therefore △DOA\cong △BOC$ (AAS congruency)

$DA=BC$ by CPCT.

So, in $ABCD$, since $DA\parallel CB$ and $DA=CB$

One pair of opposite sides of quadrilateral $ABCD$ is equal and parallel.

$\therefore ABCD$ is a parallelogram.