Similar Segments on Equal Bases are Equal
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Theorem
In the words of Euclid:
- Similar segments of circles on equal straight lines are equal to one another.
(The Elements: Book $\text{III}$: Proposition $24$)
Proof
Let $AEB$ and $CFD$ be similar segments of circles on equal bases $AB$ and $CD$.
Let the segment $AEB$ be applied to $CFD$ so that $A$ be placed on $C$ and $AB$ on $CD$.
Then $B$ will coincide with $D$ as $AB = CD$.
Suppose that segment $AEB$ does not coincide with segment $CFD$.
It will fall in one of three ways:
- $(1) \quad$ Inside it
- $(2) \quad$ Outside it
- $(3) \quad$ Awry, as $CGD$.
If $CFD$ falls inside or outside $AEB$, then by definition $AEB$ and $CFD$ are not similar.
But from Two Circles have at most Two Points of Intersection option $(3)$ is impossible.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $24$ of Book $\text{III}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions