Proof: By Euclid
(related to Proposition: 3.09: Condition for Point to be Center of Circle)
- For let $AB$ and $BC$ have been joined, and (then) have been cut in half at points $E$ and $F$ (respectively) [Prop. 1.10].
- And $ED$ and $FD$ being joined, let them have been drawn through to points $G$, $K$, $H$, and $L$.
- Therefore, since $AE$ is equal to $EB$, and $ED$ (is) common, the two (straight lines) $AE$, $ED$ are equal to the two (straight lines) $BE$, $ED$ (respectively).
- And the base $DA$ (is) equal to the base $DB$.
- Thus, angle $AED$ is equal to angle $BED$ [Prop. 1.8].
- Thus, angles $AED$ and $BED$ (are) each right angles [Def. 1.10] .
- Thus, $GK$ cuts $AB$ in half, and at right angles.
- And since, if some straight line in a circle cuts some (other) straight line in half, and at right angles, then the center of the circle is on the former (straight line) [Prop. 3.1 corr.] , the center of the circle is thus on $GK$.
- So, for the same (reasons), the center of circle $ABC$ is also on $HL$.
- And the straight lines $GK$ and $HL$ have no common (point) other than point $D$.
- Thus, point $D$ is the center of circle $ABC$.
- Thus, if some point is taken inside a circle, and more than two equal straight lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle.
- (Which is) the very thing it was required to show.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"