from D draw DH at right angles to AG; and because the angle DHA is a right angle, and the angle DAH less than a right angle, the side DH of the triangle DAH is less than the side DA (19. 1.). The point H, therefore, is within the circle, and therefore the straight line AG cuts the circle.

COR. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle; and

B

H

that it touches it only in one point; because, if it did meet the circle in two, it would fall within it (2.3.). Also it is evident that there can be but one straight line which touches the circle in the same point.

PROP. XVII. PROB.

To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle.

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

Find (1. 3.) the centre E of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw (11. 1.) DF at right angles to EA, join EBF, and draw AB.

AB touches the circle BCD.

Because E is the centre of the cir cles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle at E cominon to the two triangles AEB, FED; Therefore the base DF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles (4. 1.): Therefore the angle EBA is equal to the an

gle 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, touches the circle (Cor. 16. 3.): Therefore AB touches the circle; and 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 (Cor. 16. 3.)

PROP. XVIII. THEOR.

If a straight line touch a circle, the straight line drawn from the centre to the point of contact, is perpendicular to the line touching the circle.

Let the straight line DE touch the circle ABC in the point C; take the centre F, and draw the straight line FC: FC is perpendicu lar to DE.

A

For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, GCF must be (17. 1.) au acute angle; and to the greater angle the greater (19. 1.) side is opposite: 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 impossible; wherefore FG is not perpendicular to DE: In the same manner it may be shown, that no other line but FC can be perpendicular to DE; FC is therefore perpendicular to DE. Therefore, if a straight line, &c. Q. E. D. D

PROP. XIX. THEOR.

F

B

G

E

If a straight line touch 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 is 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 centre of the circle is in CA.

A

For, if not, let F be the centre, if possible, and join CF. DE touches the circle ABC, and FC is drawn from the centre to the point of contact, FC is perpendicular (18. 3.) 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 impossible: Wherefore F is not the centre of the circle ABC: In the same manner it may be shown, 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

F

Because

D

C

E

PROP. XX. THEOR.

The angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same point 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 same circumference BC for their base; the angle BDC is double of the angle BAC.

First, let D, the centre of the circle, be within the angle BAC, and join AD, and produce it to E: Because DA is equal to DB, the angle DAB is equal (5. 1.) to the angle DBA;

therefore the angles DAB, DBA together
are double of the angle DAB; but the an- .
gle BDE is equal (32. 1.) to the angles
DAB, DBA; therefore also the angle BDE
is double of the angle DAB: For the same
reason, the angle EDC is double of the
angle DAC: Therefore the whole angle B
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 demonstrated, as in the first case, that the angle EDC is double of the angle DAC, and that EDB a part of the first, is dou- E ble of DAB, a part of the other; therefore the remaining angle BDC is double of the remaining angle BAC. Therefore the angle at the centre, &c. Q. E. D.

B

E

PROP. XXI. THEOR.

The angles in the same segment of a circle are equal to one another.

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, both having the same part of the B circumference, viz. BCD, for their base; therefore the angle BFD is double (20. 3.) of the angle BAD: for the same rea

A

E

son, 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 AF to the centre, and produce to C, and join CE: Therefore the segment BADC is greater than a semicircle; 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

B

A E

D

to the whole angle BED. Wherefore the angles in the same seg ment, &e. Q. E. D.

PROP. XXII. THEOR.

The opposite angles of any quadrilateral figure described in a circle, are together equal to two right angles.

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

D

Join AC, BD. The angle CAB is equal (21. 3.) 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 angles CAB, ACB: To each of these equals add the angle ABC; and the angles ABC, ADC, are equal to the angles ABC, CAB, BCA. But ABC, CAB, BCA are equal A to two right angles (32. 1.); therefore also the angles ABC, ADC are equal to

B

two right angles: In the same manner, the angles BAD, DCB may be shown to be equal to two right angles. Therefore the opposite angles, &c. Q. E.D.

PROP. XXIII. THEOR.

Upon the same straight line, and upon the same 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 eircles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another; then, because the circles ACB, ADB, eut one another in the two points A, B, they cannot ent one another in

any other point (10. 3.): one of the seg
ments must therefore 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 similar segments of circles con-
tain (9. def. 3.) equal angles, the angle
ACB is equal to the angle ADB, the exte-
rior to the interior, which is impossible (16. 1.).
cannot be two similar segments of cireles
same line, which do not coincide. Q. E. D.

A

D

B

Therefore, there

upon the same side of the

+ PROP. XXIV. THEOR.

Similar segments of circles upon equal straight lines are equal to one another.

Let AFB, 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 segment AEB be applied to the segment CFD, so as the

point A be on C, and
the straight line AB
upon CD, the point
B shall coincide with
the point D, because
AB is equal to CD: ^

Therefore the straight line AB coinciding with CD, the segment AEB must (23. 3.) coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D.

PROP. XXV. 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 deseribe the circle of which it is the segment.

Bisect (10. 1.) AC in D, and from the point D draw (11.1.) DB at Fight angles to AC, and join AB: First, let the angles ABD, BAD be equal to one another; then the straight line BD is equal (6. 1.) to DA, and therefore to DC; and because the three straight lines DA, DB, DC, are all equal; D is the centre of the circle (9.3.): 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 semicircle. Next let the angles ABD, BAD be unequal; at the point A in the straight line AB make (23. 1.) the angle BAE equal to the angle ABD, and produce BD if neces

L