# Triangles with Equal Base and Same Height have Equal Area

## Theorem

In the words of Euclid:

(*The Elements*: Book $\text{I}$: Proposition $38$)

## Proof

Let $ABC$ and $DEF$ be triangles which have equal bases $BC$ and in the same parallels $AD$ and $BF$.

Let $AD$ be produced in the directions of $G$ and $H$.

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

Let $FH$ through $F$ be drawn parallel to $DE$.

Then each of $GBCA$ and $DEFH$ 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 $GBCA$ as $AB$ bisects it.

For the same reason, $DEF$ is half of $DEFH$.

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

$\blacksquare$

## Historical Note

This proof is Proposition $38$ 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