Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;
I say that ABCD is equal to the parallelogram EBCF.

For, since ABCD is a parallelogram,
AD is equal to BC. [I. 34]

For the same reason also
EF is equal to BC,
so that AD is also equal to EF; [C.N. 1]
and DE is common;
therefore the whole AE is equal to the whole DF. [C.N. 2]

But AB is also equal to DC; [I. 34]
therefore the two sides EA, AB are equal to the two sides FD, DC respectively,
and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]
therefore the base EB is equal to the base FC,
and the triangle EAB will be equal to the triangle FDC. [I. 4]

Let DGE be subtracted from each;
therefore the trapazium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]

Let the triangle GBC be added to each;
therefore the thole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]

Therefore, etc. Q.E.D.