Book III, is greater than FG; but FC is equal to FB; therefore FB is greater than FG, the lefs than the greater, which is impoffible. wherefore FG is not perpendicular to DE. in the fame manner it may be shewn, that no other is perpendicular to it befides FC, that is, FC is perpendi

a. 18. 3.

See N.

cular to DE. Therefore if a straight D

line, &c. Q. E. D.

IF

PROP. XIX. THEOR.

A

F

C GE

F a straight line touches a circle, and from the point of contact a ftraight line be drawn at right angles to the touching line, the center of the circle fhall be in that line.

Let the straight line DE touch the circle ABC in C, and from -C let CA be drawn at right angles to DE; the center of the circle is in CA.

For if not, let F be the center, if poffible, and join CF. Because
DE touches the circle ABC, and FC

is drawn from the center to the point
of contact, FC is perpendicular to
DE; therefore FCE is a right angle.
but ACE is also a right angle; there--

fore the angle FCE is equal to the B
angle ACE, the lefs to the greater,
which is impoffible. wherefore F is
not the center of the circle ABC. in
the same manner it may be fhewn D

that no other point which is not in

A

E

CA, is the center; that is, the center is in CA. Therefore if
Straight line, &c. Q. E. D.

PROP. XX. THEOR.

૧

THE angle at the center of a circle is double of the angle at the circumference, upon the same base, that is, upon the fame part of the circumference.

Let ABC be a circle, and BEC an angle at the center, and BAC Book III. an angle at the circumference, which have the fame circumference

BC for their bafe; the angle BEC is double

of the angle BAC.

a

A

Firft, Let E the center of the circle be within the angle BAC, and join AE, and produce it to F. 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 alfo the angle BEF is double of the angle EAB. for the fame reason, the angle FEC is double of the angle EAC. therefore the whole angle BEC is double of the whole angle BAC..

Again, Let BDC be inflected to the circumference, fo that E the center of the circle be without the angle BDC, and join DE and produce it to G. It may be demonftrated, as in the firft cafe, that the angle GEC is double of the angle GDC, and that GEB a part of the firft is double of GDB a part of the other; therefore the remaining angle BEC is double of the

G

B

B

F

A

E

2. 5. 1.

C

b. 32. .

remaining angle BDC. Therefore the angle at the center, &c. Q. E.D.

TH

PROP. XXI. THEOR.

HE angles in the fame fegment of a circle are see N. equal to one another.

Let ABCD be a circle, and BAD, BED angles in the fame fegment BAED; the angles BAD, BED are equal to one another.

Take Fthe center of the circle ABCD. and, firft, let the fegment BAED be greater than a femicircle, and join BF, FD.B and because the angle BFD is at the center, and the angle BAD at the circumfe

A

D

Book III. rence, and that they have the fame part of the circumference, viz. BCD for their bafe, therefore the angle BFD is double of the a. 20. 3. angle BAD. for the fame reason, the angle BFD is double of the angle BED. therefore the angle BAD is equal to the angle BED.

8. 32. 3.

b. 21.3.

But if the segment BAED be not greater than a femicircle, let

BAD, BED be angles in it; thefe
alfo are equal to one another. draw
AF to the center, and produce it to
C, and join CE. therefore the feg-B
ment BADC is greater than a femi-
circle; and the angles in it BAC,
BEC are equal, by the first case. for
the fame reason, the angles CAD,
CED are equal. therefore the whole
angle BAD is equal to the whole an-

A E

F

gle BED. Wherefore the angles in the fame fegment, &c. Q. E. D.

TH

PROP. XXII. THEOR..

HE oppofite angles of any quadrilateral figure defcribed in a circle, are together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle ABCD; any two of its oppofite angles are together 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,

b

BCA are equal to two right angles.
but the angle CAB is equal to the
angle CDB, because they are in the
fame fegment BADC; and the angle
ACB is equal to the angle ADB, be-

cause they are in the fame fegment A

ADCB. therefore the whole angle
ADC is equal to the angles CAB,

ACB. to each of thefe equals add

D

УВ

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 alfo the angles ABC, ADC are equal to two right angles. in the fame manner the angles BAD, DCB

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

UPON

PROP. XXIII. THEOR.

ON the fame ftraight line, and upon the fame See N. fide of it, there cannot be two fimilar fegments

of circles, not coinciding with one another.

If it be poffible, let the two fimilar fegments of circles, viz. ACB, ADB be upon the fame fide of the fame ftraight line AB, 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. one of the fegments must there

D

a. 10. 3

fore fall within the other; let ACB fall within ADB, and draw the straight line

BCD, and join CA, DA. and because the fegment ACB is fimilar to the fegment A ADB, and that fimilar fegments of circles

B

contain equal angles; the angle ACB is equal to the angle b. 11. Def. 3. ADB, the exterior to the interior, which is impoffible. There-c. 16. I. fore there cannot be two fimilar fegments of a circle upon the same side of the fame line, which do not coincide. Q. E. D.

SI

PROP. XXIV. THEOR.

IMILAR fegments of circles upon equal ftraight See N. lines, are equal to one another.

Let AEB, CFD be fimilar fegments of circles upon the equal ftraight lines AB, CD; the fegment AEB is equal to the fegment CFD.

line AB upon CD, the point B fhall coincide with the point D,

coincide with the fegWherefore fimilar feg

Book III. because AB is equal to CD. therefore the straight line AB coinciding with CD, the fegment AEB muft ment CFD, and therefore is equal to it. ments, &c. Q. E. D.

a. 23. 3.

See N.

2. IG. I. b. 11. I.

C. 6, 1.

d. 9.3

PROP. XXV. PRO B.

A Segment of a circle being given, to describe the

circle of which it is the fegment.

Let ABC be the given fegment of a circle; it is required to defcribe the circle of which it is the fegment.

C

Bifect AC in D, and from the point D draw b DB at right angles to AC, and join AB. Firft, let the angles ABD, BAD be equal to one another; then the ftraight line BD is equal to DA, and therefore to DC. and because the three ftraight lines DA, DB, DC are all equal, D is the center of the circle . from the center D, at the distance of any of the three DA, DB, DC defcribe a circle; this fhall pass thro' the other points; and the circle of which ABC is a fegment is defcribed. and because the center D is in AC,

C. 23. I.

£. 4. I.

c

the fegment ABC is a femicircle. but if the angles ABD, BAD are not equal to one another, at the point A in the straight line AB make the angle BAE equal to the angle ABD, and produce BD 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 fides 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 to the base EC. but AE was shewn to be equal to EB, wherefore alfo BE is equal to EC; and the three straight lines AE, EB, EC are therefore equal