Parallelogram on Same Base as Triangle has Twice its Area

Theorem

A parallelogram on the same base as a triangle, and in the same parallels, has twice the area of the triangle.

Proof

Let $ABCD$ be a parallelogram on the same base $BC$ as a triangle $EBC$, between the same parallels $BC$ and $AE$.

Join $AC$.

Then $\triangle ABC = \triangle EBC$ from Triangles with Same Base and Same Height have Equal Area.

But from Opposite Sides and Angles of Parallelogram are Equal, $AC$ bisects $ABCD$.

So the area of parallelogram $ABCD$ is twice the area of triangle $EBC$.

$\blacksquare$

Historical Note

This is Proposition 41 of Book I of Euclid's The Elements.