Question

Please any one can solve this...???

Answer 1

GIVEN:- BO=OD,AB=CD
TO PROOVE:- (i) Ar.(AOB)=Ar.(DOC)
CONSTRUCTION:- Draw a perpendicular bisector on AB and CD from O.

PROOF:- In triangle AOD and DOF
DO=BO.......( GIVEN)

[AB=DC....(GIVEN)]

1/2 of AB=1/2 of DC......(constructed)
DF=BD
angleBDO=angleDFO.......(90 degree we const.)
so by
R.H.S
triangle DFO congurent to DBO
Then,their area will be also equal.

we constructed the perpendicular bisector.
Area of BDO=area of DFO..........(PROOVED)
2×BDO=2×DFO
Ar.(ABO)=Ar.(DOC).........(Median divides a triangle in two equal area)


so we prooved first part that
area of ABO=area of DOC


ANS(ii)
we prooved that
2×BDO congurent to 2×DFO
ABO congurent to DOC
by CPCT(Congurent Parts of Congurent Triangle are equal)
angleBAC=angleDCA

we proved these angles equal and they are alternate angles
so,
DCIIAB

If DCIIAB and given that DC=AB
so,
ABCD is a parallelogram.........(Here we also prooved [iii] part)

We prooved that
ar.(ABO)=ar.(DOC)

SO,BY ADDING A COMMON TRIANGLE (BOC)

ABO+BOC=DOC+BOC

ar.(ABC)=ar.(DBC)


ANS(iii)


I PROOVED IN (ii) PART.





HOPE IT HELPS YOU!