Parallelograms with Equal Base and Same Height have Equal Area

Theorem
Parallelograms which have equal bases and in the same parallels are equal to one another.

Proof

 * Euclid-I-36.png

Let $$ABCD$$ and $$EBCF$$ be parallelograms with equal bases $$$$ and $$FG$$, and in the same parallels $$AF$$ and $$BC$$.

Join $$BE$$ and $$CH$$.

We have $$BC = FG$$ and $$FG = EH$$ so by Common Notion 1 we have that $$BC = EH$$.

But $$EB$$ and $$HC$$ join them.

So by Lines Joining Equal and Parallel Straight Lines, $$EB$$ and $$HC$$ are equal and parallel.

So $$EBCH$$ is a parallelogram.

So, by Parallelograms with Same Base and Same Height have Equal Area, the area of $$EBCH$$ equals the area of $$ABCD$$.

For the same reason, the area of $$EBCH$$ equals the area of $$EFGH$$.

So by Common Notion 1, the area of $$ABCD$$ equals the area of $$EFGH$$.