If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.
For let the parallelogram ABCD
have the same base
BC with the triangle
EBC, and let it be in the same parallels
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]
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 EBC.
Therefore, etc. Q.E.D.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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