Problem. An arc (ABC) of a circle being given, it is required to describe the whole circle.
Sol.—Take any three points A, B, C in the arc. Join AB, BC. Bisect AB in D, and BC in E. Erect DF, EF at right angles to AB, BC; then F, the point of intersection, will be the centre of the circle.
Dem.—Because DF bisects the chord AB and is perpendicular to it, it passes through the centre [i., Cor. 1]. In like manner EF passes through the centre. Hence the point F must be the centre; and the circle described from F as centre, with FA as radius, will be the circle required.