PROP. XXI. THEOREM.

In equal circles, equal angles ftand upon equal arcs, whether they be at the centres or circumferences; and if the arcs be equal; the angles will be equal.

Let ABC, DEF be two equal circles, having the angles AGB, DHE, at their centres, equal to each other, as also the angles ACB, DFE; at their circumferences; then will the arc AKB be equal to the arc DLE.

For, join the points AB, DE: then, fince the circles are equal to each other (by Hyp.), their radii and circumferences will also be equal (III. 5.)

And, fince the two fides AG, GB of the triangle ABG, are equal to the two fides DH, HE of the triangle DEH, and the angle AGB to the angle DHE (by Hyp.); their bases AB, DE will likewife be equal (I. 4.)

The chord AB, therefore, being equal to the chord DE, and the angle ACB to the angle DEE (by Hyp.), the are BCA will also be equal to the arc EF D (Cor. III. 20.)

But fince the whole circumference of the circle ABC, is equal to the whole circumference of the circle DEF, and the arc BCA to the arc EFD, the arc AKB will also be equal to the arc DLE.

H 2

Again,

Again, let the arc AKB be equal to the arc DLE; then will the angle AGB be equal to the angle DHE, and the angle ACB to the angle DFE.

For, if AGB be not equal to DHE, one of them must be greater than the other; let AGB be the greater; and make the angle AGK equal to DHE (I. 20.)

Then, fince equal angles ftand upon equal arcs (III. 21.), the arc AK will be equal to the arc DLE.

But the arc DLE is equal to the arc AKB (by Hyp.); whence the arc AK is alfo equal to the arc AKB; the less to the greater, which is impoffible.

The angle AGB, therefore, is not greater than the angle DHE; and in the fame manner it may be proved that it cannot be less; confequently they are equal to each other.

And fince angles at the centre are double to those at the circumference, the angle ACB will alfo be equal to the angle DFE. Q.E.D.

PRO P. XXII.

THEOREM.

In equal circles, equal chords fubtend equal arcs; the greater equal to the greater, and the lefs to the lefs; and if the chords be equal the arcs will be equal.

K

B

Let ABC, DEF be two equal circles, in which the chord AB is equal to the chord DE; then will the arc

ACB be equal to the arc DFE, and the arc AKB to the

arc DLE.

For, find G, H, the centres of the circles (III. 1.), and join GA, GB, HD and HE.

Then, fince the circles are equal to each other (by Hyp.) their radii and circumferences will alfo be equal (III. 5.)

And, fince the fides AG, GB are equal to the fides DH, HE, and the base AB to the bafe DE (by Hyp.), the angle AGB will also be equal to the angle DHE (I. 22.)

But equal angles, at the centres of equal circles, ftand upon equal arcs (III. 21.); therefore the arc AKB is equal to the arc DLE,

And fince the whole circumference of the circle ABC is equal to the whole circumference of the circle DEF, and the arc AKB to the arc DLE, the arc ACB will also be equal to the arc DFE.

Again, let ABC, DEF be two equal circles, of which the arc AKB is equal to the arc DLE; then will the chord AB be equal to the chord DE.

For let G, H be the centres of the circles, found as before; and join AG, GB, DH and HE,

Then, fince the circles are equal to each other (by Hyp.), the radii AG, GB will be equal to the radii pH, HE (III. 5.)

And because the arc AKB is equal to the arc DLE (by Hyp.), the angles AGB, DHE, at the centres, will be equal (III. 21.)

But, fince the two fides AG, GB are equal to the two fides DH, HE, and the angle AGB to the angle DHE, the bafe AB will also be equal to the bafe DE (1.4.) Q.E.D.

To bifect a given arc, that is, to divide it into two equal parts.

Let ADB be the given arc; it is required to divide it into two equal parts.

Draw the right line AB, which bifect in c (I. 10.); and, from the point C, erect the perpendicular CD (I. 11.); then will the arc ADB be bifected in the point D, as was required.

For, join the points AD, DB; then, fince the twọ fides AC, CD, of the triangle ADC, are equal to the two fides BC, CD of the triangle BDC, and the angle ACD to the angle BCD (I. 8.), the bafe AD will be equal to the bafe DB (I.4.)

And, because DC, or DC produced, paffes through the centre of the circle (III. 1 Cor.), the fegments ADE, DBṛ will be each of them lefs than a femicircle,

But equal chords are fubtended by equal arcs, the greater equal to the greater, and the lefs to the lefs (III, 22.); whence the chord AD being equal to the chord DB, the arc AED will be equal to the arc DFB. Q.E.I.

SCHOLIUM. An arc of a circle cannot, in general, be trifected, or divided into three equal parts, by any known method, which is purely geometrical.

The angle formed by a tangent to a circle and a chord drawn from the point of contact, is equal to the angle in the alternate fegment.

Let Be be a tangent to the circle AFDE, and aɛ a ehord drawn from the point of contact; then will the angle CAE be equal to the angle AFE in the alternate fegment,

For draw the diameter AD (III. 1.) and join the points F, D:

Then, because BC is a tangent to the circle, and AD is a line drawn through the centre, from the point of contact, the angle DAC will be a right angle (III. 12.)

And, because AFD is a femi-circle, the angle DFA will also be a right angle (III, 16.), and equal to the angle

DAC.

But fince all angles in the fame fegment of a circle are equal to each other (III. 15.), the angle DFE will be equal to the angle DAE.

If, therefore, from the equal angles DAC, DFA, there be taken the equal angles DAE, DFE, the remaining angle CAE will be equal to the remaining angle AFE. Q.E.D.