 (diagram 1) Let there be parallelograms, ABGD, EBGZ on the same base BG and in the same parallels, AZ, BG. I say that ABGD is equal to parallelogram EBGZ. (diagram 2) For since ABGD is a parallelogram, AD is equal to BG. For the same reasons, in fact, EZ is also equal to BG. Thus, AD is also equal to EZ. (diagram 3) And DE is common. Therefore, a whole, AE, is equal to a whole, DZ. (diagram 4) But AB is also equal to DG. In fact, two, EA, AB are respectively equal to two, ZD, DG. (diagram 5) And an angle, that by ADG is equal to that by EAB, the outside to the outside. (diagram 6) Therefore, a base, EB, is equal to a base, AG, and triangle EAB will be equal to triangle DZG. (diagram 7) Let a common, DHE, be taken away. Therefore, a remainder, trapezoid ABHD, is equal to a remainder, trapezoid EHGZ. (diagram 8) Let a common be added, triangle HBG. Therefore, a whole, parallelogram ABGD, is equal to a whole, parallelogram EBGZ. (diagram 9) Therefore, parallelograms that are on the same base and in the same parallels are equal to one another, just what it was required to show.