locilre

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.
The 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]

But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it [I. 34]; so that the parallelogram ABCD is also double of the triangle of EBC. Therefore etc. Q.E.D.

Sir. Thomas L. Heath, Euclid.