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THE radius of a circle is the straight line drawn from the centre to the circumference.

I.

A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it.

II.

Circles are said to touch one another, which meet, but do not cut one another.

III.

Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal.

IV.

And the straight line on which the greater perpendicular falls, is said to be farther from the centre.

B.

An arch of a circle is any part of the circumference.

V.

A segment of a circle is the figure contained by a straight line, and the arch which cuts it off.

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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.

C

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 radii of the same circle: 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 (7. 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,

F

E

G

B

D

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.

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 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.

A

C

D

E B

F

Take any point in AB as E; find D the centre of the circle ABC; join AD, DB and DE, and let DE meet the circumference in F. 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, 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: Now to the greater angle the greater side is opposite (19. 1.); DB is therefore greater than DE: but BD is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D.

PROP. III. THEOR.

If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and if it cut it at right angles, it will 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.

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: but the base EA is equal to the base
EB; therefore the angle AFE is equal (8.
1.) to the angle BFE. And when a straight
line standing upon another makes the ad-
jacent angles equal to one another, each of
them is a right (7. Def. 1.) angle: There-
fore each of the angles AFE, BFE is a
right angle; wherefore the straight line CD, A
drawn through the centre bisecting AB,
which does not pass through the centre,
cuts AB at right angles.

C

E

B

F

D

Again, let CD cut AB at right angles; CD also bisects AB, that is, AF is equal to FB.

The same construction being made, because the radii EA, EB 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; now 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 be 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 A.
through the centre, take (1. 3.) F the
centre of the circle, and join EF: and
because FE, a straight line through the
centre, bisects another AC, which does
not pass through the centre, it must cut
it at right (3. 3) angles; wherefore FEA
is a right angle. Again, because the

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B

straight line FE bisects the straight line BD, which does not pass through the centre, it must 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. É. D.

PROP. V. THEOR.

If two circles cut one another, they cannot have the same centre

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

C

G

F

D

E

For, if it be possible, let E be their centre: join EC, and draw any straight line EFG meeting the circles 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 equal to ÉG, 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.

A

PROP. VI. THEOR.

B

If two circles touch one another internally, they cannot have

the same centre..

Let the two circles ABC, CDE, touch one another internally in

the point C: they have not the same centre.

For, if they have, let it be F; join FC, and draw any straight line

FEB meeting the circles 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 FE: but CF was shown to be equal to FB; therefore 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.

A

PROP. VII. THEOR.

C

F

E

B

D

If any point be taken in the diameter of a circle which is not the centre, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line passing through the centre is always greater than one more remote from it: And from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line.

Let ABCD be a circle, and AD its diameter, in which let any pointF be taken which is not the centre: let the centre be E; of all the

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