option ii is the correct answer

Given,ABCD is a quadrilateral in which AB = CD its diagonals AC and BD intersect at O such that OB=OD.

To show:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.Construction,

DE ⊥ AC andBF ⊥ AC aredrawn.

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (each 90°)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

Therefore, ΔDOE ≅ΔBOF

(by AAScongruence rule)

Thus, DE = BF (By CPCT) — (i)

also, ar(ΔDOE) = ar(ΔBOF) ........(ii)

(Two Congruenttriangles have equal areas)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (each 90°)

CD = AB (Given)

DE = BF (From i)

Therefore,ΔDEC ≅ΔBFA

byy RHScongruence rule)

Thus, ar(ΔDEC) = ar(ΔBFA) ........(iii)

(TwoCongruent triangles have equal areas)

Adding (ii) and (iii),

ar(ΔDOE)+ ar(ΔDEC) = ar(ΔBOF)+ ar(ΔBFA)

ar (DOC) = ar (AOB)

(ii)ar(ΔDOC) = ar(ΔAOB)

⇒ar(ΔDOC)+ ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)

(Addingar(ΔOCB) on both sides)

ar(ΔDCB)= ar(ΔACB)

(iii)From part (ii) (ΔDCB) & (ΔACB) have all areas and have the same base BC.So, (ΔDCB) & (ΔACB) must lie between the same parallels.

DA || BC — (iv)

∠FBO= ∠EDO.....(v)

(ΔDOE ≅ ΔBOF )

∠FBA=∠EDC.....(vi)

(ΔDEC ≅ ΔBFA )

On adding eq v & vi

∠ABD=∠CDB

Therfore,DC||AB......(vii)

From eqiv & vii, We get DA||CB & DC||AB

Hence,ABCD is parallelogram .

ii) is the correct answer