Looking very carefully at Figure G.58, we can see several things. First, we see that EFCD is a rectangle. Second, we can see that ADE is congruent to BCF. We can think of cutting off ADE from the left side of the parallelogram and pasting it on the right side as BCF. In this way we change the figure from a parallelogram into a rectangle but we do not change the area. Third, the length of AB is b, which is also the length of EF. Therefore, the area of parallelogram ABCD is equal to the area of rectangle EFCD, which is bh. We have thus established the following: