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In Fig. 9.25, diagonals $ A C $ and $ B D $ of quadrilateral $ A B C D $ intersect at $ O $ such that $ O B=O D $. If $ \mathrm{AB}=\mathrm{CD}, $ then show that:
$ \operatorname{ar}(\mathrm{DOC})=\operatorname{ar}(\mathrm{AOB}) $
"
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 prove that $ar (DOC) = ar (AOB)$
Solution:
Draw $DP \perp AC$ and $BQ \perp AC$.
In $\vartriangle DOP$ and $\vartriangle BOQ$,
$\angle DPO = \angle BQO=90^o$
$\angle DOP = \angle BOQ$ (Vertically opposite angles)
$OD = OB$ (Given)
$\vartriangle DOP \cong \vartriangle BOQ$ (By AAS congruence)
Therefore,
$DP = BQ$----(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,
$\vartriangle CDP = \vartriangle ABQ$ (By RHS congruence)
This implies,
$ar(CDP) = ar(ABQ)$ (Since, area of congruent triangles is equal)
Adding equations (ii) and (iii), we get,
$ar(DOP) + ar(CDP) = ar(BOQ) + ar(ABQ)$
$ar (DOC) = ar (AOB)$
Hence proved .
Updated on: 10-Oct-2022
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