Dem.—If possible, let ACB, ADB, be two similar segments constructed on the same side of AB. Take any point D in the inner one. Join AD, and produce it to meet the outer one in C. Join BC, BD. Then since the segments are similar, the angle ADB is equal to ACB (Def. x.), which is impossible [I. xvi.]. Hence two similar segments not coinciding cannot be described on the same chord and on the same side of it. PROP. XXIV.—Theorem. Similar segments of circles (AEB, CFD) on equal chords (AB, CD) are equal to one another. Dem.—Since the lines are equal, if AB be applied to CD, so that the point A will coincide with C, and the line AB with CD, the point B shall coincide with D; and because the segments are similar, they must coincide [xxiii.]. Hence they are equal. 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. Dem.—1. Suppose the angles at the centres to be given equal. Now because the circles are equal their radii are equal (Def. i.). Therefore the two triangles AOB, DHE have the sides AO, OB in one respectively equal to the sides DH, HE in the other, and the angle AOB equal to DHE (hyp.). Therefore [I. iv.] the base AB is equal to DE. Again, since the angles ACB, DFE are [xx.] the halves of the equal angles AOB, DHE, they are equal [I. Axiom vii.]. Therefore (Def. x.) the segments ACB, DFE are similar, and their chords AB, DE have been proved equal; therefore [xxiv.] the segments are equal. And taking these equals from the whole circles, which are equal (hyp.), the remaining segments AGB, DKE are equal. Hence the arcs AGB, DKE are equal.