Ar(AOD) = Ar(BOC) ...(1)
If you draw the diagram, you will see triangles ΔACD and ΔBDC.
Ar(ACD) = Ar(AOD) + Ar(DOC)
Ar(BDC) = Ar(BOC) + Ar(DOC)
=> Ar(BDC) = Ar(AOD) + Ar(DOC) (From 1)
∴ Ar(ACD) = Ar(BDC) = Ar(AOD) + Ar(DOC)
(Theorem: When two triangles have same base and lie between the same two lines, and have same area, those two lines are parallel.)
∵ Two sides are parallel,
∴ ABCD is a trapezium.