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

 * Euclid-I-41.png

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 $$EBC$$.