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