Parallelograms with Same Base and Same Height have Equal Area

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

Proof

 * Euclid-I-35-Proof.png

Let $$ABCD$$ and $$EBCF$$ be parallelograms on the same base and in the same parallels $$AF$$ and $$BC$$.

As $$ABCD$$ is a parallelogram then $$AD = BC$$ from Opposite Sides and Angles of Parallelogram are Equal.

For the same reason, $$EF = BC$$.

So by Common Notion 1 we have that $$AD = EF$$.

Also, $$DE$$ is common, so the whole of $$AE$$ equals the whole of $$DF$$, by Common Notion 2.

But from Opposite Sides and Angles of Parallelogram are Equal, we have $$AB = DC$$.

From Parallel Implies Equal Corresponding Angles, $$\angle FDC = \angle EAB$$.

So by Triangle Side-Angle-Side Equality, $$EB = FC$$ and so $$\triangle EAB = \triangle FDC$$.

Subtract $$\triangle DGE$$ from each.

Then, by Common Notion 3, the trapezium $$ABGD$$ equals the trapezium $$EGCF$$.

Now we add $$\triangle GBC$$ to both.

Then, by Common Notion 2, $$ABCD = EBCF$$.