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contained by a straight line and
tained by two straight lines drawn
upon the circumference intercepted
ed by two straight lines drawn from
are those in which the an-
To find the centre of a given circle.
Draw within it any straight line AB, and bisect a it in D;
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
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
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
Take a E the centre of the circle, and join EA, EB. Then,
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;
But let CD cut AB at right angles ;
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
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:
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
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.
Book III. For, if it be possible, let E be their centre: Join EC, and
draw any straight line EFG
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