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Diagonal AC and BD of quadilateral ABCD intersect at O such that OB=OD. if AB=CD then show that 1 area DOC= area AOB (can we solve it by SAS) 2 ar DCB=ar ACB(can we solve it by same base and between same parallel).?
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Diagonal AC and BD of quadilateral ABCD intersect at O such that OB=OD...
Given- OB=OD & AB=CD
TO proove- 1.ar(AOB) =ar(DOC) 
                    2.ar (DCB) =ar(ACB) 

1 PROOF   In triangle AOB & DOC
                OB=OC (GIVEN) 
               angle ODC=angle OBA (alternate interior angle) 
                AB=DC(GIVEN) 
so triangle AOB congruent to triangle DOC
HENCE  ar(AOB)  = ar(DOC) 

2 triangle DCB and triangle ACB stands on same base BC and lies between same parallel BC and AD. 
hence ar(DCB) = ar(ACB) 

           HENCE PROOVED
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Diagonal AC and BD of quadilateral ABCD intersect at O such that OB=OD. if AB=CD then show that 1 area DOC= area AOB (can we solve it by SAS) 2 ar DCB=ar ACB(can we solve it by same base and between same parallel).?
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