a 18. 3.

See Note.

PROP. XIX. THEOR.

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 shall 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 centre of the circle is in CA.

For, if not, let F be the centre, if possible, and join CF:

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

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 part of the circumference.

Let ABC be a circle, and BEC an angle at the centre, and Book III. BAC an angle at the circumference, which have the same cir

cumference BC for their base; 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: 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

b

F

A

E

a 5. 1.

b 32.1.

C

reason, the angle FEC is double of 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 demonstrated, as in the first case, 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 remaining angle BEC is double of the remaining angle BDC. Therefore the angle at the centre, &c. Q. E. D.

G

A

PROP. XXI. THEOR.

THE angles in the same segment of a circle are See Note. equal to one another.

Book III. the circumference, viz. BCD, for their base; therefore the angle BFD is double of the angle BAD: for the same reason, the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle BED.

a 20. 3.

A E

D

F

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 it to C, and join CE: therefore the seg- B ment 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 ČAD, CED are equal: therefore the whole angle BAD is equal to the

C

whole angle BED. Wherefore the angles in the same segment, &c. Q. E. D.

a 32.1.

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.

Join AC, BD; and because the three angles of every triangle are equal a 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

b 21.3. is equal 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; therefore
the angles ABC, CAB, BCA are

A

D

7 в

equal to the angles ABC, ADC: but ABC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles: in the same manner, the angles

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

PROP. XXIII. THEOR.

UPON the same straight line, and upon the same See Note. side of it, there cannot be two similar segments of circles, not coinciding with one another.

D

a 10. 3.

If it be possible, let the two similar segments of circles, viz. ACB, ABD be upon the same side of the same straight 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 a: one of the segments must therefore fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join CA, DA: and because the segment ACB is similar to the segment A

B

ADB, and that similar segments of circles contain b equal an- b 11.def 3 gles; the angle ACB is equal to the angle ADB, the exterior to

the interior, which is impossible. Therefore, there cannot be c 16. 1. two similar segments of a circle upon the same side of the same line, which do not coincide. Q. E. D.

PROP. XXIV. THEOR.

SIMILAR segments of circles upon equal straight See Note. lines, are equal to one another.

Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD; the segment AEB is equal to the seg. ment CFD.

For, if the segment AEB beapplied to the seg

ment CFD, so as

the point A be on

C,and the straight

A

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

Book III. because AB is equal to CD: therefore the straight line AB coinciding with CD, the segment AEB must a coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D.

a 23. 3.

See Note.

a 10. 1. b 11. 1. c 6. 1.

d 9.3.

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 describe the circle of which it is the segment.

b

c

Bisecta AC in D, and from the point D draw ↳ DB at right angles to AC, and join AB: first, let the angles ABD, BAD, be equal to one another; then the straight line BD is equal 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 d: 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 se

e 23. 1.

f 4. 1.

c

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 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 angle CDE, for each of them is a right angle; therefore the base AE is equal f to the base EC: but AE was shown to be equal to EB, wherefore also BE is equal to EC: and the three straight lines AE