Similar Segments on Equal Bases are Equal

Proof

 * Euclid-III-24.png

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.