Proof. Superpose segment ABC upon segment ADC, by folding it over about AC as an axis.

Then, arc ABC will coincide with arc ADC; for otherwise there would be points of the circumference unequally distant from the centre.

Hence, segments ABC and ADC coincide throughout, and are equal.

Therefore, AC bisects the O, and its circumference.

PROP. II. THEOREM.

153. A straight line cannot intersect a circumference at more than two points.

Given

the centre of a O, and MN any str. line.

To Prove that MN cannot intersect the circumference at more than two points.

Proof. If possible, let MN intersect the circumference at three points, A, B, and C'; draw radii OA, OB, and OC. Then, OA = OB = OC.

(§ 143)

We should then have three equal str. lines drawn from a point to a str. line.

But this is impossible; for it follows from § 49 that not more than two equal str. lines can be drawn from a point to a str. line.

Hence, MN cannot intersect the circumference at more than two points.

Х

Ex. 1. What is the locus of points at a given distance from a given point?

PROP. III. THEOREM.

154. In equal circles, or in the same circle, equal central angles intercept equal arcs on the circumference.

Given ACB and A'C'B' equal central s of equal © AMB and A'M'B', respectively.

To Prove
Proof.

arc AB arc A'B'.

Superpose sector ABC upon sector A'B'C' in such

a way that ≤ C shall coincide with its equal C'.

AC A'C' and BC:

B'C'.

Now, (§ 145) Whence, point A will fall at A', and point B at B'. Then, arc AB will coincide with arc A'B'; for all points of either are equally distant from the centre.

.. arc AB arc A'B'.

PROP. IV. THEOREM.

155. (Converse of Prop. III.) In equal circles, or in the same circle, equal arcs are intercepted by equal central angles.

Given ACB and A'C'B' central of equal © AMB and A'M'B', respectively, and arc AB = arc A'B'.

To Prove

Proof. Since the

are equal, we may superpose O AMB

upon © A'M'B' in such a way that point A shall fall at A',

and centre-Cat C".

Then since arc AB = arc A'B', point B will fall at B'.

Whence, radii AC and BC will coincide with radii A'C and B'C', respectively.

Hence, C will coincide with C.

156. Sch. In equal circles, or in the same circle,

(Ax. 3)

1. The greater of two central angles intercepts the greater arc on the circumference.

2. The greater of two arcs is intercepted by the greater central angle.

PROP. V. THEOREM.

157. In equal circles, or in the same circle, equal chords subtend equal arcs.

Proof. Draw radii AC, BC, A'C', and B'C'.

Then in AABC and A'B'C', by hyp.,

Х

PROP. VI. THEOREM.

158. (Converse of Prop. V.) In equal circles, or in the same circle, equal arcs are subtended by equal chords.

(Fig. of Prop. V.)

Given, in equal © AMB and A'M'B', arc AB = arc A'B' ; and chords AB and A'B'.

Ex. 2. If two circumferences intersect each other, the distance between their centres is greater than the difference of their radii.

PROP. VII. THEOREM.

(§ 62.)

159. In equal circles, or in the same circle, the greater of two arcs is subtended by the greater chord; each arc being less than a semi-circumference.

AMB and A'M'B', arc AB> arc A'B', a semi-circumference, and chords AB and

Given, in equal

each arc being

A'B'.

To Prove

chord AB> chord A'B'.

Proof. Draw radii AC, BC, A'C', and B'C'.

Then in ▲ ABC and A'B'C',

And since, by hyp., arc AB> arc A'B', we have

(?)

(§. 156, 2)

(§ 91)

PROP. VIII. THEOREM.

160. (Converse of Prop. VII.) In equal circles, or in the same circle, the greater of two chords subtends the greater arc; each arc being less than a semi-circumference.

(Fig. of Prop. VII.)

(<C><C, by § 92; the theorem follows by § 156, 1.)

161. Sch. If each arc is greater than a semi-circumference, the greater arc is subtended by the less chord; and conversely the greater chord subtends the less arc.

PROP. IX. THEOREM.

162. The diameter perpendicular to a chord bisects the chord and its subtended arcs.

OA = OB.

Given, in O ABD, diameter CDL chord AB.

To Prove that CD bisects chord AB, and arcs ACB and ADB.

Proof. Let O be the centre of the O, and draw radii OA and OB.

Then,

Hence, ▲ OAB is isosceles.

(?)

Hence, CD bisects AB, and arcs ACB and ADB.

(§ 154)

(§ 31, 2)

(?)