Parallelograms with Equal Base and Same Height have Equal Area

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{I}$: Proposition $36$)

Proof

Let $ABCD$ and $EFGH$ be parallelograms with equal bases $BC$ and $FG$, and in the same parallels $AH$ and $BG$.

Join $BE$ and $CH$.

We have $BC = FG$ and $FG = EH$

So by Common Notion 1 we have $BC = EH$.

But $EB$ and $HC$ join them.

So by Lines Joining Equal and Parallel Straight Lines are Parallel, $EB$ 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 $ABCD$.

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

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

$\blacksquare$

Historical Note

This proof is Proposition $36$ 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