This demonstration may be stated as follows:—Since the chords are equal, they are congruent; and therefore the segments, being similar, must be congruent.
PROP. XXV.—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.
PROP. XXVI.—Theorem.
The four Propositions xxvi.–xxix. are so like in their enunciations that students frequently substitute one for another. The following scheme will assist in remembering them:—
| In Proposition | xxvi. are | given | angles | =, to | prove | arcs | =, |
| ,, | xxvii. | ,, | arcs | =, | ,, | angles | =, |
| ,, | xxviii. | ,, | chords | =, | ,, | arcs | =, |
| ,, | xxix. | ,, | arcs | =, | ,, | chords | =; |
so that Proposition xxvii. is the converse of xxvi., and xxix. of xxviii.
In equal circles (ACB, DFE), equal angles at the centres (AOB, DHE) or at the circumferences (ACB, DFE) stand upon equal arcs.