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  • In Fig. 9.25, 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)

Q: 6     In Fig. \small 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that. If \small AB=CD, then show that:

             (i)  \small ar(DOC)=ar(AOB)

            [Hint: From D and B, draw perpendiculars to AC.]

            

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We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE \perp AC and FB \perpAC

In \DeltaDEO and \Delta BFO
\angleDOE = \angleBOF [vertically opposite angle]
\angleOED = \angleBFO [each 90^0]
OB = OD [given]

Therefore, by AAS congruency
\DeltaDEO \cong  \Delta BFO
\Rightarrow DE = FB [by CPCT]

and ar( \DeltaDEO) =  ar(\DeltaBFO) ............(i)

Now, In  \DeltaDEC and  \DeltaABF
\angleDEC = \angleBFA [ each 90^0]
DE = FB
DC = BA [given]
So, by RHS congruency 
\DeltaDEC \cong  \Delta BFA
\angle1 = \angle2 [by CPCT]
and, ar( \DeltaDEC) =  ar(\DeltaBFA).....(ii)

By adding equation(i) and (ii), we get
\small ar(DOC)=ar(AOB)
Hence proved.

 

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manish

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