Let us draw DN perpendicular to AC and BM perpendicular to AC

(i) In ΔDON and ΔBOM,

DNO = BMO (By construction)

DON = BOM (Vertically opposite angles)

OD = OB (Given)

By AAS congruence rule,

ΔDON ΔBOM

DN = BM (i)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM) (ii)

In ΔDNC and ΔBMA,

DNC = BMA (By construction)

CD = AB (Given)

DN = BM [Using equation (i)]

ΔDNC ΔBMA (RHS congruence rule)

Area (ΔDNC) = Area (ΔBMA) (iii)

Let us draw DN AC and BM AC

On adding equations (ii) and (iii), we obtain

Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)

Therefore,

Area (ΔDOC) = Area (ΔAOB)

(ii) Area (ΔDOC) = Area (ΔAOB)

Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)

(Adding Area (ΔOCB) to both sides)

Area (ΔDCB) = Area (ΔACB)

(iii) Area (ΔDCB) = Area (ΔACB)

If two triangles have the same base and equal areas, then these will lie between the same parallels

DA || CB (iv)

In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of opposite sides is parallel (DA || CB)

Therefore, ABCD is a parallelogram