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

 * Euclid-I-38.png

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, $$\triangle ABC = \triangle DEF$$.