SECTION II.

CHORDS.

THEOREM 4.

In the same circle, or in equal circles, equal arcs are subtended by equal chords; and of two unequal minor arcs, the greater is subtended by the greater chord*

Part. En. Let AB, CD be equal arcs of the same or of equal circles;

it is required to prove that the chord AB is equal to the chord CD..

Proof. Let E, F be the centres of the circles: join AE, BE, CF, DF.

Then because the arc AB is equal to the arc CD; (Hyp.) therefore the angle AEB is equal to the angle CFD; (111. 3.) and because in the two triangles AEB, CFD, the two sides

*Euclid, III. 29.

AE, EB are equal to the two CF, FD, and the contained angle AEB is equal to the contained angle CFD; therefore the base AB is equal to the base CD.

(I. 5.)

Part. En. Again, let the minor arc AB be greater than the minor arc CD; it is required to prove that the chord AB is greater than the chord CD.

Proof. Because the arc AB is greater than the arc CD, therefore the angle AEB is greater than the angle CFD;

(III. 3.) and because in the triangles AEB, CFD, the two sides AE, EB are equal to the two CF, FD, but the contained angle AEB is greater than the contained angle CFD; therefore the base AB is greater than the base CD. (1. 14.)

COR. In the same circle, or in equal circles, of two unequal major arcs the greater is subtended by the less chord.

Obs. Since AB, CD are minor arcs, the angles AEB, CFD are less than two right angles. This has been assumed in the proof in treating AÉB, CFD as triangles.

THEOREM 5.

In the same circle, or in equal circles, equal chords subtend equal major and equal minor arcs; and of two unequal chords the greater subtends the greater minor arc and the less major arc*.

* Euclid, III. 28.

Part. En. Let AB, CD be equal chords in the same

or equal circles, it is required to prove that the arcs AB,

CD are equal.

Proof. For if the minor arcs AB, CD were unequal, one of them would be the greater;

and therefore the chord AB would be unequal to the chord CD.

But it is not; for the chords are equal.

Therefore the minor arcs AB, CD are also equal.

(III. 4.) (Hyp.)

Again, let the chord AB be greater than the chord CD;

it is required to prove that the minor arc AGB is greater than the minor arc CHD.

Proof. For the minor arc AGB is either greater than, equal to, or less than the minor arc CHD;

But the minor arc AGB is not equal to the minor arc CHD,

for then the chord AB would be equal to the chord CD. (III. 4.)

But it is not;

Nor is the minor arc AGB less than the minor arc CHD, for then the chord AB would be less than the chord CD.

(III. 4.)

But it is not;

therefore the minor arc AGB is greater than the minor arc CHD,

and therefore also the major arc AIB is less than the major arc CKD.

Obs. As before, the rule of conversion applies to the groups of theorems enunciated in Theorem 4 and Cor., and their converses form Theorem 5.

THEOREM 6.

The straight line drawn from the centre of a circle to the middle point of a chord is perpendicular to the chord.

Part. En. Let the straight line OC be drawn from the centre O of a circle to the middle point C of a chord AB; it is required to prove that OC is

perpendicular to AB.

Proof. Join OA, OB.

Then because in the triangles OAC, OBC, the three sides of the one are respectively equal to the three sides of the other;

therefore the angles OCA, OCB op

posite to the equal sides OA, OB are equal (1. 15), and are therefore right angles,

that is, OC is perpendicular to AB*.

THEOREM 7.

The straight line drawn from the centre of a circle perpendicular to a chord bisects the chord↑.

* Euclid, III. 3, Part 1.

Euclid, III. 3, Part 2.

Part. En. Let the straight line OC drawn from the centre of a circle be perpendicular to the chord AB; it is required to prove that OC bisects AB. ·

Proof. Because in the right-angled triangles ACO, BCO,

the hypotenuse AO is equal to the hypotenuse BO, and the side OC is common,

therefore the triangles are equal in all respects, (1. 20, Cor.) and therefore AC is equal to CB, and OC bisects AB.

THEOREM 8.

The straight line drawn perpendicular to a chord of a circle through its middle point passes through the centre of the circle*.

Part. En. Let AB be a chord of a circle, bisected in C, and let CO be drawn at right angles to AB;

it is required to prove that CO passes through the centre.

Proof. Because CO bisects AB at right angles; (Hyp.) therefore CO is the locus of points equidistant from A and B (p. 72). But the centre of the circle is equidistant from A and B:

therefore CO passes through the

centre.

COR.

B

The locus of the cen

tres of all circles that pass through

two given points is the straight

line that bisects at right angles the line joining those points.