Triangles with Same Base and Same Height have Equal Area

Theorem

In the words of Euclid:

Triangles which are on the same base and in the same parallels are equal to one another.

(The Elements: Book $\text{I}$: Proposition $37$)

Proof

Let $ABC$ and $DBC$ be triangles which are on the same base $BC$ and in the same parallels $AD$ and $BC$.

Let $AD$ be produced in the directions of $E$ and $F$.

Let $BE$ through $B$ be drawn parallel to $CA$.

Let $CF$ through $C$ be drawn parallel to $BD$.

Then each of $EBCA$ and $DBCF$ are parallelograms, and by Parallelograms with Same Base and Same Height have Equal Area they have equal areas.

From Opposite Sides and Angles of Parallelogram are Equal, $ABC$ is half of $EBCA$ as $AB$ bisects it.

For the same reason, $DBC$ is half of $DBCF$.

But by Common Notion 1, $\triangle ABC = \triangle DBC$.

$\blacksquare$

Historical Note

This proof is Proposition $37$ 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