Triangles with Equal Base and Same Height have Equal Area

Theorem

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

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 is Proposition 38 of Book I of Euclid's The Elements.