Dem.—Let O, H be the centres (see last fig.). Join AO, OB, DH, HE; then because the circles are equal, the angles AOB, DHE at the centres, which stand on the equal arcs AGB, DKE, are equal [xxvii.]. Again, because the triangles AOB, DHE have the two sides AO, OB in one respectively equal to the two sides DH, HE in the other, and the angle AOB equal to the angle DHE, the base AB of one is equal to the base DE of the other. Observation.—Since the two circles in the four last Propositions are equal, they are congruent figures, and the truth of the Propositions is evident by superposition. PROP. XXX.—Problem. To bisect a given arc ACB. Sol.—Draw the chord AB; bisect it in D; erect DC at right angles to AB, meeting the arc in C; then the arc is bisected in C. Dem.—Join AC, BC. Then the triangles ADC, BDC have the side AD equal to DB (const.), and DC common to both, and the angle ADC equal to the angle BDC, each being right. Hence the base AC is equal to the base BC. Therefore [xxviii.] the arc AC is equal to the arc BC. Hence the arc AB is bisected in C. Exercises. 1. ABCD is a semicircle whose diameter is AD; the chord BC produced meets AD produced in E: prove that if CE is equal to the radius, the arc AB is equal to three times CD. 2. The internal and the external bisectors of the vertical angle of a triangle inscribed in a circle meet the circumference again in points equidistant from the extremities of the base. 3. If from A, one of the points of intersection of two given circles, two chords ACD, AC′D′ be drawn, cutting the circles in the points C, D; C′, D′, the triangles BCD, BC′D′, formed by joining these to the second point B of intersection of the circles, are equiangular. 4. If the vertical angle ACB of a triangle inscribed in a circle be bisected by a line CD, which meets the circle again in D, and from D perpendiculars DE, DF be drawn to the sides, one of which must be produced: prove that EA is equal to BF, and hence show that CE is equal to half the sum of AC, BC. PROP. XXXI.—Theorem. In a circle—(1). The angle in a semicircle is a right angle. (2). The angle in a segment greater than a semicircle is an acute angle. (3). The angle in a segment less than a semicircle is an obtuse angle. Dem.—(1). Let AB be the diameter, C any point in the semicircle. Join AC, CB. The angle ACB is a right angle. For let O be the centre. Join OC, and produce AC to F. Then because AO is equal to OC, the angle ACO is equal to the angle OAC. In like manner, the angle OCB is equal to CBO. Hence the angle ACB is equal to the sum of the two angles BAC, CBA; but [I. xxxii.] the angle FCB is equal to the sum of the two interior angles BAC, CBA of the triangle ABC. Hence the angle ACB is equal to its adjacent angle FCB, and therefore it is a right angle [I. Def. xiii.].