Construction of Circle from Segment/Proof 2

Proof

 * Euclid-III-25d.png

Choose any point $C$ on the circumference.

Bisect $AC$ at $D$ and $BC$ at $E$ and construct a perpendicular $DF$ and $EF$ from each through the point of bisection.

The point of intersection $F$ is the center of the required circle.

$AFC$ and $CFB$ are isosceles triangles and so $AF, CF$ and $BF$ are all equal.

The result follows from Condition for Point to be Center of Circle.