Triangles with Equal Base and Same Height have Equal Area
Theorem
In the words of Euclid:
(The Elements: Book $\text{I}$: Proposition $38$)
Proof
Let $ABC$ and $DEF$ be triangles which have equal bases $BC$ and in the same parallels $AD$ and $BF$.
Let $AD$ be produced in the directions of $G$ and $H$.
Let $BG$ through $B$ be drawn parallel to $CA$.
Let $FH$ through $F$ be drawn parallel to $DE$.
Then each of $GBCA$ and $DEFH$ are parallelograms, and by Parallelograms with Same Base and Same Height have Equal Area they have equal areas.
From Opposite Sides and Angles of Parallelogram are Equal, $ABC$ is half of $GBCA$ as $AB$ bisects it.
For the same reason, $DEF$ is half of $DEFH$.
But by Common Notion 1, $\triangle ABC = \triangle DEF$.
$\blacksquare$
Historical Note
This proof is Proposition $38$ 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