If a point is taken within a circle, and more than two equal straight lines fall from the point on the circle, then the point taken is the center of the circle.

Let *D* a point within a circle *ABC,* and from *D* let more than two equal straight lines, namely *DA* and *DB* and *DC,* fall on the circle *ABC.*

I say that the point *D* is the center of the circle *ABC.*

Join *AB* and *BC,* and bisect them at the points *E* and *F.* Join *ED* and *FD,* and draw them through to the points *G, K, H,* and *L.*

Then, since *AE* equals *EB,* and *ED* is common, the two sides *AE* and *ED* equal the two sides *BE* and *ED,* and the base *DA* equals the base *DB,* therefore the angle *AED* equals the angle *BED.*

Therefore the angles *AED* and *BED* are each right. Therefore *GK* cuts *AB* into two equal parts and at right angles.

And since, if in a circle a straight line cuts a straight line into two equal parts and at right angles, the center of the circle is on the cutting straight line, therefore the center of the circle is on *GK.*

For the same reason the center of the circle *ABC* is also on *HL.*

And the straight lines *GK* and *HL* have no other point common but the point *D,* therefore the point *D* is the center of the circle *ABC.*

Therefore *if a point is taken within a circle, and more than two equal straight lines fall from the point on the circle, then the point taken is the center of the circle.*

Q.E.D.

This proposition is used in III.25.