Triangles with Same Base and Same Height have Equal Area

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

Proof

 * Euclid-I-37.png

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, $$\triangle ABC = \triangle DBC$$.