For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE;

I say that the parallelogram ABCD is double of the triangle BEC.

For let AC be joined.

Then the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [[I. 37](elem.1.37)]