angle DAE of the triangle ADE is equal to the angle HBK of

and the angle DAE is equal to the angle HBK ; therefore the side DE is equal to the side HK.

I. 5.

Next, let the minor arc DE be greater than the minor arc HK:

then shall the chord DE be greater than the chord HK.

Because the minor arc DE is greater than the minor arc HK:

therefore the angle DAE of the triangle ADE is greater than the

angle HBK of the triangle BHK;

hence in the triangles ADE, BHK, the side AD is equal to the side BH, the side AE is equal to the side BK,

and the angle DAE is greater than the angle HBK; therefore the base DE is greater than the base HK.

III. 5.

I. 16.

Q.E.D.

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

Ex. 23. If in equal circles one arc is double another, the chord which subtends the first arc is less than the double of the chord which subtends the second arc.

THEOR. 7. 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.

Let DE, HK be chords of the equal circles DEF, HKL : then according as the chord DE is greater than, equal to, or less than, the chord HK, so shall the minor arc DE be greater than equal to, or less than, the minor arc HK, and the major arc DE, be less than, equal to, or greater than, the major arc HK.

For in the preceding Theorem it has been shewn that if the minor arc DE is greater than the minor arc HK, then the chord DE is greater than the chord HK; if the minor arc DE is equal to the minor arc HK, then the chord DE is equal to the chord HK; if the minor arc DE is less than the minor arc HK, then the chord DE is less than the chord HK.

Now of these hypotheses one must be true, and of the conclusions no two can be true at the same time; hence, by the Rule of Conversion, according as the chord DE is greater than, equal to, or less than, the chord HK, so is the minor arc DE greater than, equal to, or less than, the minor arc HK ; and, therefore, the whole circumferences DEF, HKL being equal the major arc DE less than, equal to, or greater than, the major arc HK.

Q.E.D.

PROB. I. To bisect a given arc.

Let ABC be the given arc: it is required to bisect it.

B

D

Join AC.

Draw the straight line DE bisecting the line AC at right angles.

Let DE meet the arc ABC at B;

I. Prob. 4.

then shall the arc ABC be bisected at B.

Because DE is the locus of points equidistant from A and C ;

Loci, iii

therefore the chord AB is equal to the chord BC; therefore the arcs AB and BC, being both minor arcs, are equal;

III. 7.

that is, the arc ABC is bisected at B.

Q.E.F.

THEOR. 8. The straight line drawn from the centre to the middle point of a chord is perpendicular to the chord.

Let O be the centre of the circle ABC, D the middle point of a chord AB:

B

then shall the straight line OD be perpendicular to the chord AB. Join OA, OB.

I. 18.

Q.E.D.

therefore the angle ODA is equal to the angle ODB;

therefore OD is perpendicular to AB.

THEOR. 9. The straight line drawn from the centre perpendicular to a chord bisects the chord.

Let O be the centre of the circle ABC, OD the perpendicular from O to the chord AB:

8

then shall AB be bisected at D.

Join OA, OB.

Then in the triangles OAD, OBD,

the side OA is equal to the side OB,

the side OD is common to both,

III. Def. 1.

therefore AD is equal to BD;

and the angle ODA, ODB are right angles;

I. 20, Cor. 1.

Q.E.D.

that is, AB is bisected at D.

*Ex. 24. The locus of the middle points of parallel chords in a circle is a diameter of the circle.

*Ex. 25. A straight line passing through a common point of two intersecting circles has one extremity on each circumference. Shew that the projection on it of the line joining the centres is half the first straight line.

THEOR. IO. The straight line drawn perpendicular to a chord through its middle point passes through the centre.

Let AB be a chord of the circle ABC, CE the straight line drawn perpendicular to AB through D, its middle point :

D

B

then shall CE pass through the centre.