Any two shapes are said to be equivalent if they have equal areas. These terminology is quite obsolete and, nowadays, is virtually out of use.

Let ABCD be a parallelogram, BD a diagonal and M any point on it. Join MA and MC. Draw AE and CF to BD. As the triangles ABD, CBD are halves of the same , they are equal in all their parts; therefore AE = CF. The triangles AMB and CMB have the same and equal , therefore they are equivalent. By the same method of reasoning the triangles AMD, CMB are shown to be .

Another solution would be to note that for any two positions of M, say, M and M' triangles MM'A and MM'C are . For one position of M, i.e. when M is the point of intersection of the diagonals AC and BD, we have two pairs of equal and therefore equivalent pairs: AMB, CMD and AMD, CMB. For another point M', triangles AM'B, CM'D are obtained one by adding and another by subtracting triangles MM'A and MM'C from triangles AMB, CMD. And similarly, for triangles AMD, CMB. Thus changing M maintains equivalence in the two triangle pairs.