Question

diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD if AB= CD then show that : 1) area (DOC )= area(AOB) 2) area ( DCB)= area (ACB)
3) DA ll to CB or ABCD is a parallelogram

Answer 1

Heya,

GIVEN:-
Diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD. If AB=CD

TO PROVE:-
1) area (DOC )= area(AOB)
2) area ( DCB)= area (ACB)
3) DA ll to CB or ABCD is a parallelogram

CONSTRUCTION:-
From D and B, Draw perpendicular bisector to AC

PROOF:-
1) ΔDMO, ΔBNO

∠1 = ∠2 => A
∠3 = ∠4 => A
OB = OD => S

Therefore, ΔDMO ≅ ΔBNO [AAS ≅] - {1}
=> DM=BN [cpct]

Similarly, We can prove ΔDMC ≅ ΔABN [RHS ≅] - {2}

Using {1}&{2}
ΔDOC≅ ΔAOB

Therefore,
2) ar (DOC)=ar (AOB)
[add ar (COB) on both sides]

So,
ar (DCB)=ar (ACB)

3) Since,
ΔDOC≅ ΔAOB
=> ∠CDO = ∠ABO[cpct]
But these are Alternative interior angles.

So,
AB||CD and AB=CD

Hence ABCD is a parallelogram

Hope my answer helps you :)

Regards,
Shobana

Answer 2

Answer: hope this may help you

Step-by-step explanation: