8. A secant is a straight line cutting a circle.

9. A sector is the figure contained by two radii and the arc between them.

10. The angle in a segment, is the angle contained by two straight lines drawn from any point of the circumference to the extremities of the line which forms the base of the segment.

D

SECANT

SECTOR

E

BAC is an angle in the segment BDAC, and standing on the arc BEC.

11. An inscribed figure is one which has all its angles in the circumference of the circle; and the circle is said to circumscribe the figure.

12. An inscribed circle is one which touches all the sides of the figure; and the figure is said to circumscribe the circle, or be described about it.

Proposition 1.

Problem.-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 (I. 10) it in D; from the point D, draw (I. 11) DC at right angles to AB, and produce it to E, and bisect ČE in F. The point F is the centre of the circle.

For if not, let, if possible, G be the centre, and join GA, GD, GB. Then, because DA is equal to DB, and AG to GB, being radii, and DG common to the two triangles ADG, BDG, therefore the angle ADG is equal (I. 4) to the angle GDB. Therefore (I. Def. 8) the angle GDB is a right angle. But

F G

E

D

B

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.

Corollary. 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.

Proposition 2.

Theorem.-If any two points be taken in the circumference of a circle, the straight line which joins them will fall within the circle.

Let ABC be the circle, and A, B any two points in the circumference; the straight line drawn from A to B will fall within the circle.

C

Take E, any point in AB; join DE, and let it meet the circle in F. Now, because DA is equal to DB, DAB is equal (I. 13) to DBA; and because DEB is exterior, it is greater (I. 18) than DAE, the opposite interior angle; hence DEB is greater than DBE, and therefore DB is greater than (I. 21) DE. Hence DF is greater than DE, and E is within the circle; and the same may be proved of any point of AB.

Proposition 3.

B

F

Theorem.-If a straight line through the centre of a circle bisect a line not passing through the centre, it will cut it at right angles; and if it cut it at right angles, it will bisect it.

Let ABC be a circle, and let CD, a diameter, bisect AB, a chord, in the point F; it cuts it at right angles.

Take (III. 1) E, the centre of the

circle, and join EA, EB. Then, because in the triangles AEF, BEF, AE=EB, being radii, AF=FB (Hyp.), and EF common, therefore (I. 4) AFE=BFE; and (I. Def. 8) they are right angles.

Again, let CD cut AB at right angles; then will AF = FB.

Because AE= BE, then (I. 13) EAB

E

B

D

= EBA; also the angles at F are right angles, and EF is common to both triangles; therefore (I. 24) AF = FB.

Corollary. The perpendicular through the middle of a chord, and terminated both ways by the circumference, is a diameter, and its middle point is the centre.

Proposition 4.

Theorem.-If in a circle two straight lines, which do not pass through the centre, cut each other, they do not bisect each other.

Let ABCD be a circle, and AC, BD two straight lines in it which do not pass through the centre; they will not bisect each other.

For, if possible, let AE= EC and BE=ED.

Take (III. 1) F, the centre of the circle, and join EF; and because EF through the centre bisects AC, the angle FEC is (III. 3) a right angle.

A

F

For the same reason FED is a right angle. Hence FED is equal to FEC, the less to the greater, which is impossible; therefore AC, BD do not bisect each other.

Proposition 5.

Theorem.-If two circles cut each other, they cannot have the

same centre.

Let the two circles ABC, CDG cut each other in the points B, C; they cannot have the same centre.

For, if possible, let E be their centre. Join EC, and draw any straight line EFG, meeting the circles. Because E is the centre of ABC, EC is equal to EF; also because E is the centre of CDG, EC is equal to EG; therefore EF is equal to EG, which is impossible. Hence E is not the centre of the circles.

A

C

B

G

Proposition 6.

Theorem. If two circles touch each other internally, they cannot have the same centre.

Let ABC, CDE touch each other internally at C; they cannot have the same centre.

For, if possible, let it be F. Join FC, and draw any straight line FEB. Because F is the centre of the circles CDE, ABC, FC is equal to both FE and FB; therefore FE and FB are equal to each other, which is impossible. Hence F is not the centre of both circles.

Proposition 7.

B

Theorem.-If a point be taken in the diameter of a circle which is not the centre, and straight lines be drawn from it to the circumference,—

1. The greatest is that in which the centre is, and the other part of the diameter is the least.

2. Of the others, that which is nearer to the line through the centre, is greater than that more remote.

Let ABCD be a circle, AD its diameter, F a point in it, and E the centre. FA is the greatest line from F to the circumference, and FD the least; FB is greater than FC, and FC than FG.

take away EF, and FG>FD.

In the same way AF may be proved

D

greater, and FD less, than any other lines from F.