In trapezium ABCD the side AB is parallel to the side DC, and the diagonals AC and BD meet at O. Which statement is true whatever the lengths of the two parallel sides?

Triangles ACD and BCD stand on the same base DC and lie between the same parallels DC and AB, so they have equal areas. Take away the triangle OCD that both contain, and what is left gives area AOD = area BOC. Triangles AOB and COD are similar but equal in area only when AB = DC; O bisects the diagonals only in a parallelogram; and AOD and BOC have equal areas without being congruent in general, since they need not even have the same shape.