# 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