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Book III.

VI.
A segment of a circle is the figure

contained 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.”

VIII.
An angle in a segment is the angle con-

tained by two straight lines drawn
from any point in the circumference
of the segment, to the extremities
of the straight line, which is the
base of the segment.

IX.
And an angle is said to insist or stand

upon the circumference intercepted
between the straight lines that con-
tain the angle.

X.
The sector of a circle is the figure contain-

ed by two straight lines drawn from
the centre, and the circumference be-
tween them.

XI.
Similar segments of a circle

are those in which the an-
gles are equal, or which
contain equal angles.

1

PROP. I.

PROB.

See N.

centre.

To find the centre of a given circle.
Let ABC be the given circle ; it is required to find its

Draw within it any straight line AB, and bisect a it in D;
from the point D draw DC at right angles to AB, and pro-
duce it to E, and bisect CE in F: The point F is the centre
of the circle ABC.

a 10. 1. b 11. 1.

For, if it be not, let, if possible, G be the centre, and join Book III. 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

С equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn

G from the centre G*: Therefore the

F angle ADG is equal to the angle

c 8. 1. GDB: But when a straight line

A

B standing upon another straight line

ID makes the adjacent angles equal to one another, each of the angles is a

E right angle :d Therefore the angle

d 10. Def. 1. 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.

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 must fall within the circle.

a l. 3.

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

С A to B must fall within the circle.

For, if it do not, let it fall, if possible, without, as AEB; find a D the centre of the circle ABC, and join AD, DB,

D and produce DF, any straight line meeting the circumference AB, to E: Then because DA is equal to DB, the

F angle DAB is equalbto the angle DBA; A E B and because AE, a side of the triangle

b 5. 1.

* 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.

.

Book III. DAE, is produced to B, the angle DEB is greater than the

angle DAE; but DAE is equal to the angle DBE; therefore c 16. 1.

the angle DEB is greater than the angle DBE: but to the d 19. 1. greater angle the greater side is opposite;d 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 impos-
sible: Therefore the straight line drawn from A to B does not
fall without the circle. In the same manner, it may be de-
monstrated, 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.

a l. 3.

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 also cuts it at
right angles.

Take a E the centre of the circle, and join EA, EB. Then,
because AF is equal to FB, and FE common to the two tri-
angles 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 to the angle BFE.
But when a straight line standing upon
another, makes the adjacent angles

equal to one another, each of them is E c 10. Def. I. a right angle: Therefore each of the

angles AFE, BFE is a right angle;
wherefore the straight line CD, drawn

F 'B
through the centre bisecting another
AB that does not pass through the

D
centre, cuts the same at right angles.

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

The same construction being made, because EA, EB from d 5. I. the centre are equal to one another, the angle EAF is equal

to the angle EBF; and the right angle AFE is equal to the
right angle BFE: Therefore, in the two triangles, EAF, EBF,

b 8. 1.

there are two angles in the one equal to two angles in the other, Book III. 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; ° AF therefore is equal to FB. Wherefore, “ if a e 26. 1.

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 the 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 a F the

a l. 3. centre of the circle, and join EF:

F

D and because FE, a straight line

A through the centre, bisects another AC, which does not pass through the centre, it must cut it at right

с

b 3. 3. angles; wherefore FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, it must cut it at right 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 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.

G

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Book III. For, if it be possible, let E be their centre: Join EC, and

draw any straight line EFG
meeting the circles in F and

С
G; and because E is the centre
of the circle ABC, CE is equal
to EF: Again, because E is A
the centre of the circle CDG,

D
CE is equal to EG: But CE
was shewn to be equal to EF;
therefore EF is equal to EG,
the less to the greater, which is
impossible: Therefore E is not

B
the centre of the circles ABC,
CDG. Wherefore, “if two circles," &c. Q. E. D.

PROP. VI. THEOR.

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

С
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: And CF

B
was shewn equal to FB; there- A F E
fore FE is equal to FB, the less
to the greater, which is impossible:

D
Wherefore F is not the centre of
the circles ABC, CDE. There-
fore, “ if two circles,&c. Q. E. D.

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