If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle.
Proof
Let F be the point and FA, FB, FC three equal lines to the
circle. Join AB, BC; bisect them at G, H (I.10). Join
FG, FH. In △FAG and △FBG: FA=FB given,
AG=GB by construction, FG common; by I.8 the triangles are
congruent, so ∠FGA=∠FGB, and by I.13 both are right.
Similarly FH⊥BC. By III.3 (rewriting it as: the
perpendicular at the midpoint of a chord passes through the centre),
both FG produced and FH produced pass through the centre. Their
intersection F is therefore the centre.