Book III. Therefore FC 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 fhewn, that no other is perpendicular to it befides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

A

D

B GE

@ 18.

I'

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

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

A

F

For, if not, let F be the centre, if poffible, and join CF : Becaufe DE touches the circle ABC, and FC is drawn from the centre to the point of contact, FC is perpendicular to DE; therefore FCE is a right angle: But ACE is alfo a right angle; therefore the angle FCE is equal to the angle ACE, the lefs to B the greater, which is impoffible: Wherefore F is not the centre of the circle ABC: In the fame manner, it may be fhewn, that no other point. which is not in CA, is the centre;

D

E

that is, the centre is in CA. Therefore, if a ftraight line, &c. Q. E. D.

See No

THE
'HE angle at the centre of a circle is double of the
angle at the circumference, upon the fame base,
that is, upon the fame part of the circumference.

Let

Let ABC be a circle, and BEC an angle at the centre, and Book III. BAC an angle at the circumference, which have the fame circumference BC for their bafe; the angle 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 F: Becaufe 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 B EAB, EBA; therefore alfo the angle BEF is double of the angle EAB: For the fame reason, the angle FEC is double of

A

E

a 5. I.

b 32. I.

C

F

the angle EAC: Therefore the whole angle BEC is double of the whole angle BAC.

Again, Let E the centre of the circle be without the angle BDC, and join DE and produce it to G. It may be demonftrated, as in the first cafe, that the angle GEC is double 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

D

E

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

THE

to one another.

a

Book III. the circumference, viz. BCD, for their bafe; therefore the an. gle BFD is double of the 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.

a 20. 3.

But, if the fegment BAED be not greater than a femicircle,

let BAD, BED be angles in it; these

alfo are equal to one another: Draw
AF to the centre, 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 firft cafe: For
the fame reafon, becaufe CBED is
greater than a femicircle, the angles.
CAD, CED are equal: Therefore
the whole angle BAD is equal to the

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

a 32. I.

bai. 3.

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

angles.

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

a

D

Join AC, BD; and becaufe 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 fame fegment BADC; and the angle ACB is equal to the angle ADB, because they are in the fame fegment ADCB: Therefore the whole angle ADC is equal to the A angles CAB, ACB: To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA are e

B

qual 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,

BAD, DCB may be fhewn to be equal to two right angles. Book 111. Therefore, the oppofite angles, &c. Q. E. D.

PON the fame ftraight line, and upon the fame fide See N.

UPO 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 straight line
AB, not coinciding with one another: Then, because the cir-
cle 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 therefore fall within
the other; let ACB fall within ADB,
and draw the ftraight line BCD, and
join CA, DA: And because the feg- A
ment ACB is fimilar to the fegment

B

a IO. 3.

ADB, and that fimilar fegments of circles contain equal an- b11. def. 3. gles; the angle ACB is equal to the angle ADB, the exterior

to the interior, which is impoffible. Therefore, there cannot c 16. 1. be two fimilar fegments of a circle upon the fame fide of the

fame line, which do not coincide. Q. E. D.

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.

AA

For, if the fegment AEB be applied to the fegment CFD, fo as the point A be on C, and A the ftraight line

B C

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

E 4

caufe

a

Book III. caufe AB is equal to CD: Therefore the ftraight line AB coinciding with CD, the fegment AEB muft fegment CFD, and therefore is equal to it. fegments, &c. Q. E. D.

a 23. 3.

coincide with the Wherefore fimilar

Sce N.

a 10. 1. b II. I.

c 6.1.

d 9. 3.

A

PROP. XXV. PROB.

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.

Bifect AC in D, and from the point D draw b DB at right angles to AC, and join AB: First, 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 centre of the circle: From the centre D, at the distance of any of the three DA, DB, DC, defcribe a circle; this fhall pass through the other points; and the circle of which ABC is a fegment is described : And because the centre D is in AC, the fegment ABC is a fe

€ 23. I.

f 4.1.

micircle: 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, if neceffary, to E, and join EC: And because the angle ABE is equal to the angle BAE, the ftraight 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 bafe EC: But AE was fhewn to be equal to EB, wherefore alfo BE is equal to EC: And the three ftraight lines AE,

EB,