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equal + to the angle ACE, the less to the greater; which + 1 Ax. is impossible; therefore F is not the centre of the circle ABC. In the same manner it may be shewn, that no other point which is not in CA, is the centre ; that is, the centre is in CA. Therefore, if a straight line, &c.

Q. E. D.

PROPOSITION XX.

Tusor.— The angle at the centre of a circle is double of the See N.

angle at the circumference upon the same base, that is, upon the same part of the circumference.

* 5. I.

* 32. 1.

B В

Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base : the angle BEC shall be double of the angle BAC.

Join AE, and produce it to F. First, let the centre of the circle be within the angle BAC. Because EA is equal to EB, the angle EAB is equal * to the angle EBA ; therefore the angles EAB, EBA are double of the angle EAB : but the angle BEF is equal * to the angles EAB, EBA ; therefore also the angle BEF is double of the angle EAB : for the same reason, the angle FEC is double of the angle EAC; therefore the whole angle BEC is double of the whole angle BAC.

Again, let the centre of the circle be without the angle BAC: it may be demonstrated,

A as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part F of the first, is double of FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC. Therefore, the angle at the centre, &c. Q. E. D.

B В

PROPOSITION XXI.

THEOR.The angles in the same segment of a circle are See N.

equal to one another.

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

† 1. 3.

20.3.

+7 Ax.

First, let the segment BAED be greater than a semicircle. Take † F, the centre of the circle ABCD, and join BF, FD : and because the angle BFD is at the centre, and the angle BAD at the circumference, and

E that they have the same part of the circumference, viz. BCD for their base, therefore the angle BFD is double * of the angle BAD: for B В the same reason, the angle BFD is double of the angle BED; therefore the angle BAD is equal † to the angle BED.

Next, let the segment BAED be not greater than a semicircle. Draw AF to the centre, and produce it to C, and join CE: therefore the segment BADC A E is greater than a semicircle; and the angles in В, it, BAC, BEC are equal, by the first case : for

F 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 angles in the same segment, &c.

Q. E, D.

+ 2 Ax.

PROPOSITION XXII.

THEOR.- The opposite angles of any quadrilateral figure

inscribed in a circle, are together equal to two right angles.

• 32. 1.

* 21. 3.

Let ABCD be a quadraliteral figure in the circle ABCD: any two of its opposite angles shall together be equal to two right angles.

Join AC, BD: and because the three angles of every triangle 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 is equal * to the angle CDB, because they are in the same segment CDAB; and the angle ACB is equal to the angle ADB, because they are in the same

B segment ADCB; therefore the two angles CAB, ACB are together equal + to the whole angle ADC: to each of these equals add the angle ABC; therefore the three angles ABC, CAB, BCA are equal † to the two angles ABC, ADC: but ABC, CAB, BCA are equal to two right

† 2 Ax.

+ 2 Ax.

angles ; therefore also the angles ABC, ADC are equal + to + 1 Ax. two right angles. In the same manner the angles BAD, DCB may be shewn to be equal to two right angles. Therefore, the opposite angles, &c.

Q. E. D.

PROPOSITION XXIII.

Theor.Upon the same straight line, and upon the same side Sce N.

of it, there cannot be two similar segments of circles, not coinciding with one another.

If it be possible, upon the same straight line AB, and upon the same side of it, let there be two similar segments of circles, ACB, ADB, not coinciding with one another. Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other * point :

• 10. 3.

А therefore one of the segments must fall within the other: let ACB fall within ADB ; draw the straight line BCD, and join CA, DA. And because the segment ACB is similar + to the segment ADB, and that similar segments of † Hyp. circles contain * equal angles, therefore the angle ACB is '11 Def. 3. equal to the angle ADBI, the exterior to the interior ; which is impossible. Therefore, there cannot be two similar seg- . 16. 1. ments of circles upon the same side of the same line, which do not coincide. Q. E. D.

PROPOSITION XXIV.
THEOR.-Similar segments of circles upon equal straight lines, See N.

are equal to one another.

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

For if the segment AEB be ap-
plied to the segment CFD, so that
the point A may be on C, and the straight line AB upon CD,
the point B shall coincide with the point D, because AB is equal

That is, ACB, the exterior angle of the triangle ACD, is proved to be equal to the interior and opposite angle ADC* ; which is impossible.

• 16, 1.

. 23. 3.

to CD: therefore the straight line AB coinciding with CD, the segment AEB must * coincide with the segment CFD, and therefore is equal + to it. Wherefore, similar segments, &c.

78 Ax.

Q. E. D.

PROPOSITION XXV.

See N.

ProB.-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. • 10. 1. Bisect * AC in D, and from the point D, draw * DB at right # 11.1.

angles to AC, and join AB. First, let the angles ABD, BAD + See Fig. 1. be equal+ to one another; then the straight line BD is equal * • 6. 1.

B

[blocks in formation]

9.3.

2. and 3.

23. 1.

to DA, and therefore I to DC: and because the three straight lines DA, DB, DC are all equal, D is the centre of the circle *. 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 semi+ See Fig. circle. But if the angles ABD, BAD are not equal t to one

another, at the point A, in the straight line AB, make * the + See Fig. 2. 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 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 + Constr.

angle CDE, for each of them is a right t angle; therefore the • 4. I. base AE is equal * to the base EC: but AE was shewn to be

equal to EB; wherefore also BE is equal † to EC ; and therefore the three straight lines AE, EB, EC are equal to one another: wherefore * E is the centre of the circle. From the

• 6.1.

+1 Ax.

9. 3.

DC; but

| For since AC was bisected in D, therefore AD is equal
is proved to be equal to AD, therefore BD is equal t to DC.

+) Ax.

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 + See Fig. 2. 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, + See Fig. 3. which is therefore greater than a semicircle. Wherefore, a segment of a circle being given, the circle is describedo f which it is a segment. Which was to be done.

PROPOSITION XXVI.

THEOR.— In equal circles, equal angles stand upon equal cir

cumferences, whether they be at the centres or circumfer

ences.

Let ABC, DEF be equal circles, and let the angles BGC, EHF at their centres, and BAC, EDF at their circumferences, be equal to each other : the circumference BKC shall be equal to the circumference ELF.

Join BC, EF: and because the circles ABC, DEF are equal, the straight lines drawn from their centres t are equal: there- † 1 Def. 3. fore the two sides BG, GC are equal to the two EH, HF, each to each: and the angle at G is equal † to the angle at H; + Hyp. therefore the base BC is equal * to the base EF. And because * 4. I. the angle at A is equal to the angle at D, the segment BAC + Hyp. is similar * to the segment EDF; and they are upon equal

* 11 Def. 3. straight lines BC, EF: but similar segments of circles upon equal straight lines, are equal * to one

* 24. 3. another; therefore the segment

H BAC is equal to the segment

B

CES EDF: but the whole circle ABC K is equal t to the whole DEF;

+ Нур. . therefore the remaining segment BKC is equal † to the remain- † 3 Ax. ing segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c.

Q. E. D.

PROPOSITION XXVII.

THEOR.-In equal circles, the angles which stand upon equal

circumferences are equal to one another, whether they be at the centres or circumferences.

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