VI.

A segment of a circle is the figure con

tained by a straight line and the circumference it cuts off.

VII.

"The angle of a segment is that which is contained by the straight line and the circumference."

The sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.

XI.

Şimilar segments of a circle, are those in which the angles are

equal, or which contain equal angles.

PROP. I. PROB.

To find the centre of a given circle.*

Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC.

* See Note.

F

C

G

For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre G:* therefore the angle ADG is equal (8. 1.) to the angle GDB: but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle: (10. def. 1.) therefore the angle GDB is a right angle: but FDB is likewise a right angle; wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible: therefore G is not the centre of the circle ABC: in the same manner it can be shown, that no other point but F is the centre; that is, F is the centre of the circle ABC. Which was to be found

A

B

D

E

COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other.

PROP. II. THEOR.

IF Ir any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A

to B shall fall within the circle.

For, if it do not, let it fall, if possible, without, as AEB; find (1. 3.) D the centre of the circle ABC, and join AD, DB, and produce DF, any straight line meeting the circumference AB, to E: then because DA is equal to DB, the angle DAB is equal (5. 1.) to the angle DBA; and because AE, a side of the triangle DAE,

D

C

Α

E B

*N. B. Whenever the expression " straight lines from the centre," or "drawn from the centre," occurs, it is to be understood that they are drawn to the circumference.

is produced to B, the angle DEB is greater (16. 1.) than the angle DAE: but DAE is equal to the angle DBE: therefore the angle DEB is greater than the angle DBE: but to the greater angle the greater side is opposite; (19. 1.) DB is therefore greater than DE: but DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impossible: therefore the straight line drawn from A to B does not fall without the circle. In the same manner it may be demonstrated that it does not fall upon the circumference; it falls therefore within it. Wherefore, if any two points, &c. Q. E. D.

PROP. III. THEOR.

Ir a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles, and, if it cuts it at right angles, it shall bisect it.

Let ABC be a circle; and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: it cuts it also at right angles.

C

Take (1. 3.) E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB: therefore the angle AFE is equal (8. 1.) to the angle BFE: but when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right (10. def. 1.) angle: therefore each of the angles AFE, BFE is a right angle; wherefore the straight line CD, drawn through the cen- A tre bisecting another AB that does not pass through the centre, cuts the same at right angles.

E

B

D

But let CD cut AB at right angles: CD also bisects it, that is, AF is equal to FB.

The same construction being made, because EA, EB from the centre are equal to one another, the angle EAF is equal (5. 1.) to the angle EBF: and the right angle AFE is equal to the right angle BFE: therefore, in the two triangles EAF, EBF,

there are two angles in one equal to two angles in the other, and the side EF, which is opposite to one of the equal angles, in each, is common to both; therefore the other sides are equal (26. 1.): AF therefore is equal to FB. Wherefore, if a straight line, &c. Q. E. D.

PROP. IV. THEOR.

IF in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each other.

Let ABCD be a circle, and AC, BD two straight lines in it which cut one another in the point E, and do not both pass through the centre: AC, BD do not bisect one another.

For, if it is possible, let AE be equal to EC, and BE to ED if one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre; but, if neither of them pass through the centre, take (1. 3.) F the centre of the circle, and join EF: and because FE, a straight line through the centre, bi- A sects another AC which does not pass through the centre, it shall cut it at right (3. 3.) angles; wherefore FEA is

B

T

D

a right angle: again, because the straight line FE bisects the straight line BD which does not pass through the centre, it shall cut it at right (3. 3.) angles; wherefore FEB is a right angle and FEA was shown to be a right angle; therefore FEA is equal to the angle FEB, the less to the greater, which is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D.

PROP. V. THEOR.

If two circles cut one another, they shall not have the

same centre.

Let the two circles ABC, CDG cut one another in the points B, C; they have not the same centre.

D

C

For, if it be possible, let E be their centre: join EC, and draw any straight line EFG meeting them in F and G; and because E is the centre of the circle ABC, CE is equal to EF: again, because E is the centre of the circle CDG, CE is equal to EG: but CE was shown to be equal to EF; therefore EF is A equal to EG, the less to the greater, which is impossible: therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D.

PROP. VI. THEOR.

Ꮐ

F

E

B

If two circles touch one another internally, they shall not have the sanie centre.

Let the two circles ABC, CDE touch one another internally in the point C: they have not the same centre.

For, if they can, let it be F; join FC, and draw any straight line FEB meeting them in E and B;

and because F is the centre of the circle ABC, CF is equal to FB; also, because F is the centre of the circle CDE, CF is equal to FB: and CF was shown equal to FB; therefore A FE is equal to FB, the less to the greater, which is impossible: wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D.

C

B

F

E

D