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Let ABC be a circle, and BEC an angle at the centre, and Book III. BAC an angle at the circumference, which have the same circumference BC for their base; the angle

A BEC is double of the angle BAC.

First, let E the centre of the circle be within the angle BAC, and join AE, and produce it to P: because EA is equal to EB, the angle EAB is equal a to the

a 5. 1.

E angle EBA ; therefore the angles EAB, EBA are double of the angle EAB; but the angle BEF is equal b to the angles

b 32. 1 EAB, EBA ; therefore also the angle BEF B is double of the angle EAB: for the same

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F reason, the angle FEC is double of the angle EAC: therefore the whole angle BEC is double of the whole angle BAC. Again, Let E the centre of the

А circle be without the angle BDC, and join DE, and produce it to G. It

D may be demonstrated, as in the first case, that the angle GEC is double

E of the angle GDC, and that GEB a part of the first is double of GDB a part of the other; therefore the re

G maining angle BEC is double of the remaining angle BDC. Therefore, the angle at the centre, &c. Q. E. D.

B

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PROP. XXI. THEOR.

THE angles in the same segment of a circle are See N. equal to one another.

А

E

F

Let ABCD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED are equal to one another.

Take F the centre of the circle ABCD: and, first, let the segment BAED be greater than a semicircle, and join BF, FD: and because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of

B

Book III. the circumference, viz. BCD, for their base; therefore the an

gle BFB is double a of the angle BAD: for the same reason, a 20.3. the angle BFD is double of the angle BED: therefore the angle

BAD is equal to the angle BED.

But, if the segment BAED be not greater than a semicircle,
let BAD, BED be angles in it; these
also are equal to one another: draw

A E
AF to the centre, and produce it to
C, and join CE: therefore the seg-
ment BADC is greater than a semi B

D
circle; and the angles in it BAC,
BEC are equal, by the first case: for
the 'same reason, because CBED is
greater than a semicircle, the angles
CAD, CED are equal: therefore the
whole angle BAD is equal to the
whole angle BED. Wherefore, the

С angles in the same segment, &c. Q. E. D.

PROP. XXII. THEOR.

THE opposite angles of any quadrilateral figure de. scribed in a circle are together equal to two right angles.

C

Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles.

Join AC, BD; and because the three angles of every tria 32. 1. angle are equal to two right angles, the three angles of the

triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles : but the angle CAB

D b 21.3. is equal b to the angle CDB, because

they are in the same segment BADC,
and the angle ACB is equal to the
angle ADB, because they are in the
same segment ADCB: therefore the
whole angle ADC is equal to the А

B
angles CAB, ACB: to each of these
equals add the angle ABC; therefore
the angles ABC, CAB, BCA are
equal to the angles ABC, ADC: but ABC, CAB, BCA are
equal to two right angles; therefore also the angles ABC, ADC
are equal to two right angles: in the same manner, the angles

BAD, DCB may be shown to be equal to two right angles. Book III. Therefore, the opposite angles, &c. Q. E. D.

PROP. XXIII. THEOR.

UPON the same straight line, and upon the same see N. side of it, there cannot be two similar segments of circles not coinciding with one another.

If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another: then, because the circle ACB cuts the circle ADB in the two

D points A, B, they cannot cut one another in any other point a: one of the segments

a 10.3. must therefore fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join CA, DA: and because

A

B the segment ACB is similar to the segment ADB, and that similar segments of circles contain b equal an-b 11. def. gles; the angle ACB is equal to the angle ADB, the exterior to 3. the interior, which is impossibles. Therefore, there cannot be c 16. 1. two similar segments of a circle upon the same side of the same line, which do not coincide. Q. E. D.

PROP. XXIV. THEOR.

SIMILAR segments of circles upon equal straight See N. 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. For, if the seg.

E

F ment AEB be applied to the seg. ment CFD, so as

B the point A be on A

C C, and the straight line AB upon CD, the point B shall coincide with tlie point D,

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Book III. because AB is equal to CD: therefore, the straight line AB co

inciding with CD, the segment AEB must a coincide with the a 23. 3. segment CFD, and therefore is equal to it. Wherefore, similar

segments, &c. Q. E. D.

PROP. XXV. PROB.

See N.

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. a 10. 1. Bisect a AC in D, and from the point D draw b DB at right b 11. 1. angles to AC, and join AB: first, let the angles ABD, BAD c 6.1. be equal to one another; then the straight line BD is equal

to DA, and therefore to DC; and because the three straight

lines DA, DB, DC are all equal, D is the centre of the cir. d 9. 3. cled: 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 se

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micircle: but if the angles ABD, BAD are not equal to one e 23. 1. another, at the point A, in the straight line AB, make the angle

BAE equal to the angle ABD, and produce BD, if necessary, to E, and join EC: and because the angle ABE is equal to the angle BAE, the straight linc 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 f 4. 1. fto the base EC: but AE was shown to be equal to EB; where

fore also BE is equal to EC: and the three straight lines AE,

EB, EC are therefore equal to one another; wherefore d E is Book III. the centre of the circle. From the centre E, at the distance of a any of the three AE, EB, EC, describe a circle ; this shall pass d 9.3. 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. Which was to be done.

PROP. XXVI. THEOR.

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

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences : the circumference BKC is equal to the circumference 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;

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and the angle at G is equal to the angle at H; therefore the base BC is equal a to the base EF: and because the angle at A is equal a 4. 1. to the angle at D, the segment BAC is similarb to the segment b 11. def. EDF; : id they are upon equal straight lines BC, EF; but simi- 3. lar se go nts of circles upon equal straight lines are equal to one amuler: therefore the segment BAC is equal to the segment c 24.3 EDF : but the whole circle ABC is equal to the whole DEF;

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