Question

6. 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)
(ii) ar (DCB) = ar (ACB)
(111) DA | CB or ABCD is a parallelogram.
(Hint: From D and B, draw perpendiculars to AC.]
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Answer 1

Answer:

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Answer 2

ABCD is a parallelogram

Step-by-step explanation:

We have a quadrilateral ABCD whose  diagonals AC  and BD  intersect at point O

we have OB=OD ,AC=CD

let us draw a DE perpendicular to AC and BF perpendicular to AC

proof : (i) in  $\bigtriangleup DEO$ and$\bigtriangleup BFO$

DO=B0(given)

$\angle DOE$=$\angle BOF$(vertically opposite angles)

$\angle DEO=\angle BFO$(Each 90°)

$\bigtriangleup DEO \cong \bigtriangleup BFO$ (By AAS congruency)

DE= BF (by CPCT)

and ar(($\bigtriangleup  DEO$) = ar($\bigtriangleup BFO$)...(1)

now in $\bigtriangleup DEC$ and $\bigtriangleup BFA$

We have,

$\angle DEC=\angle BFA$(each 90)

DE=BF(proved above)

DC= BA(Given)

therefore,

$\bigtriangleup DEC\cong \bigtriangleup BFA$ (By RHS congruency rule)

ar($\bigtriangleup DEC$) = ar($\bigtriangleup BFA$)...(2)

and $\angle 1=\angle 2...(3)$ (by cpct)

adding (1) and (2) we have,

ar($\bigtriangleup DEO$)+ ar($\bigtriangleup DEC$)= ar($\bigtriangleup BFO$)+ar($\bigtriangleup BFA$)

ar($\bigtriangleup DOC$= ar($\bigtriangleup AOB$)

(ii)since ar($\bigtriangleup DOC$= ar($\bigtriangleup AOB$)(proved above)

adding ar($\bigtriangleup BOC$ )on both sides we have

ar($\bigtriangleup DOC$ + ar($\bigtriangleup BOC$ =ar($\bigtriangleup AOB$)+ ar($\bigtriangleup BOC$ )

ar($\bigtriangleup DCB$)= ar($\bigtriangleup ACB$)

(iii)since, $\bigtriangleup DCB$ and $\bigtriangleup ACB$ are on the same base CB and  having equal areas

they must lie between the same parallel lines CB and DA

∴CB║DA

and $\angle 1= \angle 2$ (by 3)

which are alternate interior angles

therefore,

AB║CD

hence, ABCD is a parallelogram

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