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 trapezium $ABGD$ equals the trapezium $EGCF$.
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