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For, if it be not, let, if poffible, G be the centre, and join Book III. GA, GD, GB:. Then, because DA is equal to DB, and DG

C

FG

c 8. I.

D

common to the two triangles ADG,
BDG, the two fides AD, DG are e-
qual to the two BD, DG, each to
each; and the bafe GA is equal to
the base GB, because they are drawn
from the centre G: Therefore the
angle ADG is equal to the angle
GDB: But when a ftraight line ftand-
ing upon another ftraight line makes A
the adjacent angles equal to one ano-
ther, each of the angles is a right an.
gled: 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 lefs,
which is impoffible: Therefore G is not the centre of the cir
cle ABC: In the fame manner it can be fhewn, that no other
point but F is the centre; that is, F is the centre of the cirde
ABC: Which was to be found.

E

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

d ro, def. I.

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

Let ABC be a circle, and A, B any two points in the cir

cumference; the ftraight line drawn

from A to B fhall fall within the circle.

For, if it do not, let it fall, if poffible, without, as AEB; find D the cen tre of the circle ABC, and join AD, DB, and produce DF, any ftraight line meeting the circumference AB, to E: Then because DA is equal to DB, the angle DAB is equal to the angle DBA; and because AE, a fide of the triangle

E 3

C

D

A E B

DAE,

• N. B. Whenever the expreffion " ftraight lines from the centre," or "drawn "from the centre," occurs, it is to be understood that they are drawn to the ci cumference.

a I. 3.

s 16. I.

d 19. .1

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 the angle DEB is greater than the angle DBE: But to the greater angle the greater fide is oppofite d; DB is therefore greater than DE: But DB is equal to DF; wherefore DF is greater than DE, the lefs than the greater, which is impofiible: Therefore the ftraight line drawn from A to B does not fall without the circle. In the fame 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.

a 1. 3.

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IF a ftraight line drawn through the centre of a circle bifect a straight line in it which does not pafs through the centre, it fhall cut it at right angles; and, if it cuts it at right angles, it fhall bifect it.

Let ABC be a circle; and let CD, a ftraight line drawn through the centre, bifect any straight line AB, which does not pafs through the centre, in the point F: It cuts it alfo at right angles.

C

Take E the centre of the circle, and join EA, EB: Then, becaufe AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two fides in the one equal to two fides in the other, and the base EA is equal to the bafe EB; therefore the angle AFE is equal to the Angle BFE: But when a ftraight line ftanding upon another makes the adjacent angles equal to one another, each of them is a right def. 1. angle: Therefore each of the angles

b 8. I.

AFE, BFE is a right angle; wherefore
the traight line CD, drawn through the
centre bifecting another AB that does
not pafsthrough the centre, çuts the fame
at right angles.

A

E

F

But let CD cut AB at right angles; CD alfo bifects it, that is, AF is equal to FB.

The fame conftruction being made, because EA, EB from the centre are equal to one another, the angle EAF is equal 4 to the angle EBF; and the right angle AFE is equal to the right angle BFE: Therefore, in the two triangles EAF, EBF,

there

there are two angles in one equal to two angles in the other, Book III. and the fide EF, which is oppofite to one of the equal angles in each, is common to both; therefore the other fides are equal; AF therefore is equal to FB. Wherefore, if a straight e 26. I. line, &c. Q. E. D.

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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 ftraight lines in it which cut one another in the point E, and do not both pafs through the centre: AC, BD do not bifect one another.,

For, if it is poffible, let AE be equal to EC, and BE to ED: If one of the lines pafs through the centre, it is plain that it cannot be bifected by the other which does not pass through the centre: But, if neither of them país through the centre, take F the centre of the circle, and join EF: And because FE, a ftraight line through the centre, bifects another A AC which does not pafs through the centre, it fhall cut it at right angles; B wherefore FEA is a right angle: Again, because the ftraight line FE bi

a I. 3.

F

D

E

Cb 3. 3.

fects the ftraight line BD which does not pafs through the centre, it shall cut it at right bangles; wherefore FEB is a right angle: And FEA was fhewn to be a right angle; therefore FEA is equal to the angle FEB, the lefs to the greater, which is impoffible: Therefore AC, BD do not bifect one another. Wherefore, if in a circle, &c. Q. E. D.

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

fame centre.

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

E 4

For,

Book III.

A

For, if it be poffible, let E be their centre: Join EC, and draw any ftraight line EFG meeting them in F and G: And because E is the centre of the circle ABC, CE is equal to EF: Again, becaufe E is the centre of the circle CDG, CE is equal to EG: But CE was fhewn to be equal to EF; therefore EF is equal to EG, the lefs to the greater, which is impoffible: Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D.

G

D

E

B

PROP. VI. THE OR.

IF two circles touch one another internally, they shall

not have the fame centre.

Let the two circles ABC, CDE, touch one another internally in the point C: They have not the fame 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: Al-
fo, because F is the centre of the
circle CDE, CF is equal to FE: And
CF was fhewn equal to FB; there-
fore FE is equal to FB, the lefs to
the greater, which is impoffible:
Wherefore F is not the centre of the
circles ABC, CDE. Therefore, if
two circles, &c. Q. E. D.

F

/E

D

B

PROP.

PROP. VII.

VII. THEOR.

Book III

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any point be taken in the diameter of a circle, which is not the centre, of all the ftraight 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 leaft; and, of any others, that which is nearer to the line which paffes through the centre is always greater than one more remote: And from the fame point there can be drawn only two ftraight lines that are equal to one another, one upon each fide of the shortest line.

Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre: Let the centre be E; of all the ftraight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA is the greateft, and FD, the other part of the diameter AD, is the leaft; and of the others, FB is greater than FC, and FC than FG.

Join BE, CE, GE; and because two fides of a triangle are greater than the third, BE, EF are greater than BF; but AE a 20. 1. is equal to EB; therefore AE, EF,

that is AF, is greater than BF: A

gain, because BE is equal to CE,

B

1

E

b 24. I.

G

D H

and FE common to the triangles C
BEF, CEF, the two fides BE, EF
are equal to the two CE, EF; but
the angle BEF is greater than the
angle CEF; therefore the bafe BF is
greater than the bafe FC: For the
fame reason, CF is greater than GF:
Again, because GF, FE are greater
than EG, and EG is equal to
ED; GF, FE are greater than ED: Take away the common
part FE, and the remainder GF is greater than the remainder
FD: Therefore FA is the greateft, and FD the leaft of all the
ftraight lines From F to the circumference; and BF is greater
than CF, and CF than GF.

Alfo there can be drawn only two equal ftraight lines from the point F to the circumference, one upon each fide of the

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