TROP. XXII. (THEOREM.)-The opposite angles of any quadrilateral figure (ABCD) inscribed in a circle (ABCD), are together equal to two right angles.

Join AC and BD. The angle CAB is equal to the angle CDB (III. 21), because they are in the same segment CDAB. And the angle ACB is equal to the angle ADE, because they are in the same segment ADCB. Therefore the two angles CAB and ACB are together equal (I. Ax. 2) to the whole angle ADC. To each of these equals add the angle ABC. Therefore the three angles ABC, CAB, and BCA are equal to A the two angles ABC and ADC (I. Ax. 2). the three angles ABC, CAB, and BCA are equal (I. 32) to two right angles. Therefore the two angles ABC and ADC are equal (I. Ax. 1) to two right angles. In the same manner it may be shown that the two angles BAD and DCB are equal to two right angles. Therefore the opposite angles, &c. Q. E. D.

But

B

PROP. XXIII. (THEOREM.)-Upon the same straight line (AB), and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another.

Upon the same straight line AB, and upon the same side of it, let there be, if possible, two similar segments of circles, ACB and ADB, not coinciding with one another.

Because the circumference ACB cuts the circumference ADB in the two points A and B, they cannot cnt one another (III. 10) in any other point. Therefore one of the segments must fall within the other. Let ACB fall within ADB. Draw the straight line A BCD, and join CA and DA.

D

B

Because the segment A CB is similar (Hyp.) to the segment ADB, and similar segments of circles contain (III. Def. 11) equal angles; therefore the angle ACB is equal to the angle ADB; that is, the exterior angle of the triangle ACD is equal to its interior angle, which is impossible (I. 16). Therefore upon the same straight line, and on the same side of it, there cannot be two similar segments of circles which do not coincide. Q. E. D.

PROP. XXIV. (THEOREM.)-Similar segments (AEB, CFD) of circles upon equal straight lines (AB, CD), are equal to one another, and have equal

ures.

E

F

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

BC

D

with the segment CFD, because they are similar. Therefore the segment AEB is equal (I. Ax. S) to the segment CFD, and the are AEB to the arc CFD. Wherefore similar segments, &c. Q.E. D.

PROP. XXV. (PROBLEM.)-A segment of a circle (ABC) being given, to describe the circle of which it is the segment.

Bisect AC in D (I. 10), from the point D draw DB at right angles to AC (I. 11) and join AB.

First, let the angles ABD and BAD be equal to one another.

Because the straight line BD is equal to DA (I. 6), and also (I. Ax. 1) to DC, the three straight lines DA, DB, and DC are all equal. Therefore D is the centre (III. 9) of the circle, From the centre D, at the distance of any of the three straight lines DA, DB, or DC, describe a circle, and it shall pass through the other points. Therefore the circle of which ABC is a segment is described.

Next, let the angles ABD and BAD be unequal to one

another.

At the point A, in the straight line AB, make the angle BAE equal (I. 23) to the angle ABD. Produce

BD, if necessary, to meet AE in E, and join EC,

Because the angle ABE is equal to the angle BAE, the straight line BE is equal (I. 6) to the straight line EA. Because AD is equal to DC, and DE common to the triangles ADE and CDE, the two sides AD and DE are equal to the two sides CD and DE, each to each But the angle ADE is equal to the angle CDE, for each of them. (Const.) is a right angle. Therefore the base AE is equal (I. 4) to the base EC. But AE was shown to be equal to EB. Therefore the three straight lines AE, EB, and EC are equal (I. Ax. 1) to one another, and E is the centre (III. 9) of the circle. From the centre E, at the distance of any of the three straight lines AE, EB, or EC describe a circle, and it shall pass through the other points. Therefore the circle, of which ABC is a segment, is described. Wherefore, a segment of a circle being given, the circle is described of which it is a segment. Q. E. F.

Cor.-In the first case, because the centre D is in AC, the segment ABC is a semicircle. In the second case, if the angle ABD be greater than the angle B AD, the centre E falls without the segment ABC, and it is therefore less than a semicircle. But if the angle ABD be less than BAD, the centre E falls within the segment ABC, and it is therefore greater than a semicircle.

PROP. XXVI. (THEOREM.)-In equal circles (ABC, DEF), equal angles (BGC and EHF at the centres, and BAC and EDF at the circumferences) stand upon equal arcs, whether they be at the centres or circumferences.

Join BC and EF.

Because the circles ABC and DEF are equal, the straight lines drawn from their centres (III. Def. 1) are equal; therefore the two sides BG and GC are equal to the two EH and HF, each to each. But the angle at G is equal (Hyp.) to the angle at H. Therefore the base BC is equal (I. 4) to the base EF. Because the angle at A is equal (Hyp.) to the angle at D, the segment BAC is similar (III. Def

A

11) to the segment EDF, and they are upon equal straight lines BC and EF. But similar segments of circles upon equal straight lines are equal (III. 24) to one another. Therefore the segment BAC is equal to the segment EDF, and the arc BAC to the arc EDF. But the whole circumference ABC is equal (Hyp.)to the whole circumference DEF. Therefore the remaining segment BKC is equal (I. Ax. 3) to the remaining

K

CE
L

segment ELF, and the arc BKC to the arc ELF. Wherefore, in equal circles, &c. Q. E. D.

PROP. XXVII. (THEOREM)-In equal circles (ABC and DEF), the angles (BGC and EHF at the centres, and BAC and EDF at the circumferences) which stand upon equal arcs (BC and EF) are equal to one another, whether they be at the centres or circumferences.

If the angle BGC be not equal to the angle EHF, one of them must be greater than the other. Let the angle BGC be the greater, and at the point G, in the straight line

BG, make the angle BGK equal (I. 23) to the angle EHF.

Because the angle BGK is equal to the angle EHF, and equal angles stand (III. 26) upon equal arcs; therefore the arc BK is equal to the arc EF. But the arc EF is equal

CE

(Hyp.) to the arc BC. Therefore also BK is equal (I. Ax. 1) to BC, the less to the greater, which is impossible. Wherefore the angle BGC is not unequal to the angle EHF; that is, the angle BGC is equal to the angle EHF. But the angle at A is half of the angle BGC (III. 20) and the angle at D half of the angle EHF. Therefore the angle at A is equal (I. Ax. 7) to the angle at D. Wherefore, in equal circles, &c. Q. E. D.

PROP. XXVIII. (THEOREM.)-In equal circles (ABC, DEF), equal straight lines (BC and EF) cut off equal arcs, the greater equal to the greater, and the less to the less.

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

Because the circles ABC and DEF are equal, the straight lines drawn from their centres (III. Def. 1) are equal. Therefore BK and KC

are equal to EL and LF, each to each. But the base BC is equal (Hyp.) to the base EF. Therefore the angle BKC is equal (I. 8) to the angle ELF. Because equal angles stand upon equal arcs (III. 26) the arc BGC is equal to the arc EHF.

D

K

C

E

But the whole circumference ABC is equal (Hyp) to the whole circumference EDF. Therefore, the remaining arc BAC is equal

(I. Ax. 3) to the remaining arc EDF. Therefore, in equal circles, &c. Q. E. D.

PROP. XXIX. (THEOREM.)-In equal circles (ABC, DEF) equal arcs
(BGC, EHF) are subtended by equal straight lines (BC, EF).
Take K and L (III. 1), the centres of the circles, and join BK,
KC, EL, and LF.

B

A.

D

K

L

C E

G

H

Because the arc BGC is equal to the arc EHF, the angle BKC is equal (III. 27) to the angle ELF. Because the circles ABC and DEF are equal, the straight lines from their centres are equal (III. Def. 1). Therefore BK and KC are equal to EL and LF, each to each. But the angle BKC is equal to the angle ELF. Therefore the base BC is equal (I. 4) to the base EF. Therefore, in equal circles, &c. Q. E. D.

PROP. XXX. (PROBLEM.)-To bisect a given arc (ADB); that is, to divide it into two equal parts.

Join AB, and bisect (I. 10) it at C. From the point C, draw CD at right angles (I. 11) to AB. The arc ADB is bisected at the point D.

D

Join AD and DB. Because AC is equal to CB, and CD common to the triangles ACD and BCD, the two sides AC and CD are equal to the two sides BC and CD, each to each. But the angle ACD is equal to the angle BCD, because each of them is a right angle; therefore the base AD is equal (I. 4) to the base BD. But equal straight lines cut off equal (III. 28) arcs, the greater equal to the greater, and the less to the less; and AD and DB are each of them less than a semicircle, because DC (III. 1. Cor.) passes through the centre. Therefore the arc AD is equal to the arc DB. Therefore the given arc is bisected at D. Q. E. F.

A

C

B

PROP. XXXI. (THEOREM.)-In a circle (ABCD), the angle in a semicircle (BAC) is a right angle; the angle in a segment (ABC) greater than a semicircle is less than a right angle; and the angle (ADC) in a segment less than a semicircle is greater than a right angle.

Take any point D in the arc ADC, and join AB, AD, DC, and AC. Also join AE, and produce BA to F.

F

Because BE is equal (I. Def. 15) to EA, the angle EAB is equal (I. 5) to the angle EBA. Because AE is equal to EC, the angle EAC is equal to the angle ECA. Therefore the whole angle BAC is equal (I. Ax. 2) to the two angles ABC and ACB. But FAC, the exterior angle of the triangle ABC, is equal (1.32) to the two angles ABC and ACB. Therefore the angle BAC is equal (I. Ax. 1) to B the angle FAC, and each of them (I. Def. 10) is a right angle. Therefore the angle BAC in a semicircle is a right angle. Because the two

E

angles ABC and BAC of the triangle ABC are together less than

two right angles (I. 17), and BAC has been proved to be a right angle. Therefore ABC is less than a right angle. Wherefore the angle in a segment ABC greater than a semicircle, is less than a right angle. Because ABCD is a quadrilateral figure inscribed in a circle, any two of its opposite angles are equal (III. 22) to two right angles. Therefore the angles ABC and ADC, are equal to two right angles. But ABC has been proved to be less than a right angle. Therefore the angle ADC is greater than a right angle. Wherefore the angle in a segment ADC less than a semicircle is greater than a right angle. Therefore, in a circle, &c. Q. E. D.

Cor.-From this it is manifest that if one angle of a triangle bc equal to the other two, it is a right angle: because the angle adjacent to it is equal (I. 32) to the same two; and when the adjacent angles are equal (I Def. 10) they are right angles.

PROP. XXXII. (THEOREM.)-If a straight line (EF) touch a circle (ABCD) and from the point of contact (B) a straight line (BD) be drawn cutting the circle; the angles which this straight line makes with the tangent are equal to the angles in the alternate segments of the circle. From the point B draw BA at right angles (I. 11) to EF, take any point C in the arc DCB, and join AD, DC, and CB.

F

Because the straight line EF touches the circumference of the circle ABCD at the point B, and BA is drawn at right angles to the tangent from the point of contact B, the centre of the circle is (III. 19) in BA. Therefore the angle ADB in a semicircle (III. 31) is a right angle. Because the other two angles BAD and ABD, in the triangle ADB, are equal to (I. 32) a right angle, and ABF E B is (Const.) a right angle; therefore the angle ABF is equal to the two angles BAD and ABD (I. Ax. 1). From these equals take away the common angle ABD; therefore the remaining angle DBF is equal to the remaining angle BAD (I. Ax. 3), which is in the alternate segment of the circle. Because ABCD is a quadrilateral figure in a circle, the opposite angles BAD and BCD are equal (III. 22) to two right angles. But the two angles DBF and DBE are equal (I. 13) to two right angles. Therefore the two angles DBF and DBE are equal (I. Ax. 1) to the two angles BAD and BCD. But the angle DBF has been proved equal to the angle BAD. Therefore the remaining angle DBE is equal (I. Ax. 2) to the remaining angle BCD, which is in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXXIII. (PROBLEM.)-Upon a given straight line (AB) to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle (C).

First, let the angle C be a right angle.

Bisect AB in F (I. 10), and from the centre F, at the distance FB, describe the semicircle AHB. Because the segment A H B is a semicircle, the angle AHB is a right angle (III. 31), and therefore it is equal to the given angle C.

Next, let the angle C be an oblique angle.

C

A

F