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.

In the words of Euclid:

If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

(The Elements: Book $\text{I}$: Proposition $41$)

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 proof is Proposition $41$ of Book $\text{I}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions