Proof: By Euclid

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.
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