Parallelograms with Same Base and Same Height have Equal Area

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{I}$: Proposition $35$)

Proof

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 Parallelism implies Equal Corresponding Angles:

$\angle FDC = \angle EAB$

So by Triangle Side-Angle-Side Congruence:

$EB = FC$

and so $\triangle EAB = \triangle FDC$.

Subtract $\triangle DGE$ from each.

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

Now we add $\triangle GBC$ to both.

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

$\blacksquare$

Historical Note

This proof is Proposition $35$ of Book $\text{I}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions