because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point (III. 10); one of the segments must therefore fall within the other; let ACB fall within ADB, and draw the straight line

BCD, and join CA, DA; and because the segment ACB is similar to the segment ADB, the angle ACB is equal to the angle ADB (III. Def. 8), the exterior to the interior, which is impossible (I. 16).

PROPOSITION XXIV. THEOREM.

Similar segments of circles upon equal straight lines are equal to one another.

Let AEB, CFD, be similar segments of circles upon the equal straight lines AB, CD; the segment AEB is equal to the segment CFD.

E

B

For, if the segment AEB be applied to the segment CFD, so as the point A be on C, and the straight line AB upon CD, the point B shall coincide with the point D, because AB is equal to CD; therefore the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD (III. 23), and therefore is equal to it.

A segment of a circle being given, to describe the circle of which it is the segment.

Let ABC be the given segment of a circle; it is required to describe the circle of which it is the segment.

B

Bisect AC in D (I. 10), and from the point D draw DB at right angles to AC (I. 11), and join AB; first, let the angles ABD, BAD, be equal to one another; then the straight line BD is equal to DA (I. 6), and

E

D

therefore to DC; and because the three straight lines DA, DB, DC, are all equal, D is the centre of the circle (III. 9); from the centre D, at the distance of any of the three DA,

DB, DC, describe a circle; this shall pass through the other points; and the circle, of which ABC is a segment, is described; and because the centre D is in AC, the segment ABC is a semicircle; but if the angles ABD, BAĎ, are not equal to one another, at the point A, in the straight line AB, make the angle BAE equal to the angle ABD (I. 23), and produce BD, if necessary, to E, and join EC ; and because the angle ABE is equal to the angle BAE, the straight line BE is equal to EA; and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE, are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equal to the base EC (I. 4); but AE was shown to be equal to EB, wherefore also BE is equal to EC; and the three straight lines AE, EB, EC, are therefore equal to one another; wherefore E is the centre of the circle (III. 9). From the centre E, at the distance of any of the three AE, EB, EC, describe a circle, this shall pass through the other points; and the circle, of which ABC is a segment, is described; and it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC, which therefore is less than a semicircle; but if the angle ABD be less than BAD, the centre E falls within the segment ABC, which is therefore greater than a semicircle; wherefore a segment of a circle being given, the circle is described of which it is a segment.

PROPOSITION XXVI. THEOREM.

In equal circles, equal angles stand upon equal arcs, whether they be at the centres or circumferences.

Let ABC, DEF, be equal circles, and the equal angles BGC, EHF, at their centres, and BÁC, EDF, at their circumferences; the arc BKC is equal to the arc ELF.

Join BC, EF; and because the circles ABC, DEF, are equal, the straight lines drawn from their centres are equal; therefore the two sides BG, GC, are equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF (I. 4); and

because the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF (III. Def. 8); and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines are equal to one another (III.24);

L

therefore the segment BAC is equal to the segment EDF; but the whole circle ABC is equal to the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the arc BKC to the arc ELF.

In equal circles, the angles which stand upon equal arcs are equal to one another, whether they be at the centres or circumferences.

Let the angles BGC, EHF, at the centres, and BAC, EDF, at the circumferences of the equal circles ABC, DEF, stand upon the equal arcs BC, EF; the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF.

If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to EDF (III. 20). But, if not, one of them is the

greater; let BGC be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle

B

K

EHF (I. 23); but equal angles stand upon equal arcs (III. 26), when they are at the centre; therefore the arc BK is equal to the arc EF; but EF is equal to BC; therefore also BK is equal to BC, the less to the greater, which is impossible; therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it; and the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; therefore the angle at A is equal to the angle at D.

In equal circles, equal chords cut off equal arcs, the greater equal to the greater, and the less to the less.

Let ABC, DEF, be equal circles, and BC, EF, equal chords in them, which cut off the two greater arcs BAC, EDF, and the two less BGC, EHF; the greater BAC is equal to the greater EDF; and the less BGC to the less EHF.

B

K

Take K, L, the centres of the circles (III. 1), and join BK, KC, EL, LF; and because the circles are equal, their radii are equal; therefore BK, KC, are equal to EL, LF; and the base BC is equal to the base EF; therefore the angle BKC is equal to the angle ELF (I. 8). But equal angles stand upon equal arcs (III. 26), when they are at the centres; therefore the arc BGC is equal to the arc EHF. But the whole circle ABC is equal to the whole EDF; the remaining part therefore of the circumference, namely, BAC, is equal to the remaining part EDF.

G

PROPOSITION XXIX. THEOREM.

H

In equal circles equal arcs are subtended by equal chords. Let ABC, DEF, be equal circles, and let the arcs BGC, EHF, also be equal, and join BC, EF; the chord BC is equal to the chord GF.

Take K, L, the centres of the circles (III. 1), and join BK, B KC, EL, LF; and because the

K

E

D

H

arc BGC is equal to the arc EHF, the angle BKC is equal to the angle ELF (III. 27); and because the circles ABC, DEF, are equal, their radii are equal; therefore BK, KC, are equal to EL, LF, and they contain equal angles; therefore the base BC is equal to the base EF (I. 4).

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

Let ADB be the given arc; it is required to bisect it. Join AB, and bisect it in C (I. 10); from the point C draw CD at right angles to AB, and join AD, DB; the arc ADB is bisected in the point D.

Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides AC, CD, are equal to the two BC, CD; and the angle ACD is equal to

the angle BCD, because each of them is a right angle; therefore the base AD is equal to the

base BD (I. 4). But equal chords cut off equal arcs (III. 28), the greater equal to the greater, and the less to the less; and AD, DB, are each of them less than a semicircle, because DC passes through the centre (III. 1, Cor.) ; wherefore the arc AD is equal to the arc DB; therefore the given arc is bisected in D.

In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

F

Let ABCD be a circle, of which the diameter is BC, and centre E; and draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right B angle.

D

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA (1.5); also, because AE is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to