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" />Let ABCD be the quadrilateral plot. Produce BA to meet CD drawn parallel to CA at E. Join EC. Triangles EAC and ADC lie on the same base AC and between the same parallels DE and CA. Using "Two triangles on the same base and between the same parallels are equal in area." ar(EAC) = ar(ADC) ar(ABCD) = ar(ABC) + ar(ACD) = ar(ABC) + ar(EAC) = ar(EBC) i.e., Quadrilateral ABCD = △EBC gives the explanation.