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

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Given:

In a quadrilateral ABCD diagonals \( A C \) and \( B D \)  intersect at \( O \) such that \( O B=O D \).

$AB = CD$.

To do:

We have to show that:

(i) \( \operatorname{ar}(\mathrm{DOC})=\operatorname{ar}(\mathrm{AOB}) \)
(ii) \( \operatorname{ar}(\mathrm{DCB})=\operatorname{ar}(\mathrm{ACB}) \)
(iii) DA\| \( \mathrm{CB} \) or \( \mathrm{ABCD} \) is a parallelogram.

Solution:

Draw $DP \perp AC$ and $BQ \perp AC$.

(i) In $\vartriangle DOP$ and $\vartriangle BOQ$,

$\angle DPO = \angle BQO=90^o$

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

$OD = OB$ (Given)

Therefore, by AAS congruence

$\vartriangle DOP \cong \vartriangle BOQ$

This implies,

$DP = BQ$----(i)   (CPCT)

$ar(DOP) = ar(BOQ)$----(ii)   (Since, area of congruent triangles is equal)

In $\vartriangle CDP$ and $\vartriangle ABQ$,

$\angle CPD = \angle AQB=90^o$

$CD = AB$   (Given)

$DP = BQ$   [From equation (i)]

Therefore, by RHS congruence

$\vartriangle CDP \cong \vartriangle ABQ$

This implies,

$ar(CDP) = ar(ABQ)$...........(iii)    (Since, area of congruent triangles is equal)

Adding equations (ii) and (iii), we get,

$ar(DOP) + ar(CDP) = ar(BOQ) + ar(ABQ)$

$ar (DOC) = ar (AOB)$

(ii) $ar (DOC) = ar (AOB)$

This implies,

$ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC)$       [Adding $ar (BOC)$ on both sides]

$ar(DCB) = ar (ACB)$

(iii) $\triangle DCB$ and $\triangle ACB$ have equal areas and have the same base.

This implies,

$\triangle DCB$ and $\triangle ACB$ must lie between the same parallels.

Therefore,

$DA \| CB$

$ar(DOP) = ar(BOQ)$

This implies,

$\angle OBQ = \angle PDO$…(iv)

$ar(CDP) = ar(ABQ)$

This implies,

$\angle ABQ= \angle CDP$….....(v)

Adding (iv) and (v), we get,

$\angle OBQ+\angle ABQ = \angle CDP + \angle PDO$

$\angle ABO=\angle CDO$

$\angle CDB= \angle ABD$

This implies,

$CD \| AB$

Therefore, $ABCD$ is a parallelogram.