is greater than FK: And the fquares of BH, HE are equal to Book III. the fquares of FK, KE, of which the fquare of BH is greater than the fquare of FK, because BH is greater than FK; there- 5. Def.3. fore the fquare of EH is lefs than the fquare of EK, and the ftraight line EH less than EK.

Q. E. D.

THE

Wherefore the diameter, &c.

PROP. XVI. THEOR.

'HE ftraight line drawn at right angles to the See N. diameter of a circle, from the extremity of it, falls without the circle; and no ftraight line can be drawn between that ftraight line and the circumference from the extremity, fo as not to cut the circle.

Let ABC be a circle, the centre of which is D, and the diameter AB; the ftraight line drawn at right angles to AB from its extremity A, fhall fall without the circle.

C

E

a 17. I.

Take any point E in AE, and join DE, and let it meet the circumference in C: and becaufe the two angles DAE, AED of the triangle ADE are together less than two right angles 2, and that DAE is a right angle, AED must be less than a right angle and to the greater angle the greater fide is oppofite; therefore the fide DE is greater than DA; but DC is equal to DA; therefore DE is greater than DC, and the point E is therefore without the circle: but E is any point whatever in AE; therefore AE falls without the circle.

D

And every other ftraight line drawn through the point A cuts the circle; for, let AF be any straight line, making the angle DAF less than the right angle DAE: and from the point D draw DG perpendicular to AF, meeting the circumference in H: and because DGA is a right angle, and DAG less than a right angle, DA is greater than DG: but DH is equal to DA; therefore DH is greater than DG; therefore AF falls within the circle. Wherefore, &c. Q. E. D.

b

COR. From this it is manifeft, that g the ftraight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle; and that it touches only in one point,

K

D

E

F

H

G

A

b 19. 1.

C 12. 1.

b 19. I.

BOOK III. point, because, if it did meet the circle in two, it would fall within it. Also it is evident, that there can be but one straight

d 2. 3. line which touches the circle in the same point.

a 1. 3

PROP. XVII. PROB.

O draw a ftraight line from a given point, either without or in the circumference, which fhall touch a given circle.

First, let A be a given point without the given circle BCD ; it is required to draw a ftraight line from A which shall touch the circle.

a

Find the centre E of the circle, and join AE; and from the centre E, at the distance EA, defcribe the circle AFG; from b. I. the point D draw DF at right angles to EA, and join EBF, AB; AB touches the circle BCD..

Because E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two fides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED; therefore the bafe DF is equal to the bafe AB, and the triangle FED to the triangle AEB, and the other angles to the

C

E

B

c 4. 1. other angles : Therefore the angle EBA is equal to the angle EDF: But EDF is a right angle, wherefore EBA is a right

:

angle And EB is drawn from the centre; but a straight line drawn from the extremity of a diameter, at right angles to it, dCor.16.3. touches the circle: Therefore AB touches the circle; and it is drawn from the given point A. Which was to be done.

But, if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE; DF touches the circle d.

PROP. XVIII. THEOR.

Fa ftraight line touches a circle, the ftraight line

fhall be perpendicular to the line touching the circle.

Let the ftraight line DE touch the circle ABC in the point C, take the centre F, and draw the ftraight line FC; FC is perpendicular to DE.

For,

A

b 17. I.

C 19. I.

For, if it be not, from the point F draw a FBG perpendicular Book. III. to DE; and because FGC is a right angle, GCF is an acute angle; and to the greater angle the greater fide is oppofite: a 12. I. Therefore FC is greater than FG: but FC is equal to FB; therefore FB is greater than FG, the less 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 a ftraight line, &c. Q. E. D.

PROP. XIX. THEOR.

B

Ꭰ

GE

F a ftraight line touches a circle, and from the point of contact, a ftraight 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.

Α

For, if not, let F be the centre, if poffible, and join CF: Because 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 also a right angle; therefore the angle FCE is equal to the angle ACE, the less to 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; that is, the centre is in CA. Therefore, if a straight line, &c. Q. E. D.

B

PROP. XX. THEOR.

D

C

E

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 ABC be a circle, and BDC an angle at the centre, and BAC an angle at the circumference, which have the fame circumference

K 2

a 18. 3.

Book III. cumference BC for their bafe; the angle BDC is double of the Wangle BAC.

2

First, let D the centre of the circle be within the angle BAC, and join AD, and produce it to E: Becaufe DA is equal a 5. I. to DB, the angle DAB is equal to the angle DBA; therefore the angles DAB, DBA are double of the angle DAB; b32. 1. but the angle BDE is equal to the angles DAB, DBA; therefore alfo the angle B BDE is double of the angle DAB: For

E

с

the fame reason, the angle EDC is double of the angle DAC:
Therefore the whole angle BDC is double of the whole angle
BAC.

Again, let D the centre of the circle be
without the angle BAC, and join AD,
and produce it to E. It may be demon-
strated, as in the first cafe, that the angle
EDC is double of the angle EAC, and
that EDB, a part of the firft, is double of
EAB, a part of the other; therefore the
remaining angle BDC is double of the
remaining angle BAC. Therefore the B
angle at the centre, &c. Q. E. D.

PROP. XXI, THEOR.

D

C

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

THE

equal to one another.

Let ABCD be a circle, and BAED a fegment of it; the angles in the fegment BAED are equal to one another.

Take F the centre of the circle, and
through it draw any ftraight line AC
cutting BD, and join BA, AD; and
let BED be any other angle in the feg-
ment BAED: the angle BAD is equal
to the angle BED.

Join BF, CE and because the angle
BFC is at the centre, and the angle
BAC at the circumference, and that

B

A

E

D

F

they have the fame part of the circumference, viz. BC, for a 20. 3. their bafe; therefore the angle BFC is double a of the angle BAC: For the fame reason, the angle BFC is double of the angle BEC: therefore the angle BAC is equal to the angle BEC. In like manner, it may be demonftrated, that the angle CAD is equal to the angle CED: therefore the whole angle BAD is equal to the whole BED. Wherefore, the angles, &c. Q. E. D.

PROP.

THE

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.

a

Book III.

Join AC, BD; and because the three angles of every triangle are equal to two right angles, the three angles of the a 32. 1. triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles: But the angle CAB

b

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 segment ADCB: Therefore the whole angle ADC is equal to the angles CAB, ACB: To each of thefe equals add the angle ABC; therefore

A

E

C

Է 21. 3.

B

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 opposite angles, &c. Q. E. D.

c

COR. If one of the fides DA be produced to E, the exterior angle EAB is equal to the interior oppofite angle BCD. For EAB, BAD are equal to two right angles; that is, to the two angles BCD, BAD; therefore the angle EAB is equal to BCD.

C 13. I.

U

PROP. XXIII. THEOR.

PON the fame ftraight line, and upon the fame 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 muft a 10. 3. therefore fall within the other; let ACB fall within ADB, and

draw