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For if it be not, let, if possible, G be the center, and join GA, Book III. 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 bafe GA is equal to the base GB, be

G cause they are drawn from the center G*. therefore the angle ADG is equal

c. 8. s. to the angle GDB. but when a straight A

B line standing upon another straight line,

ID makes the adjacent angles equal to one

E another, each of the angles is a right angled, therefore the angle GDB is a right angle. but FDB is d.10. Def.s. 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 center of the circle ABC. in the same manner it can be shewn, that no other point but F is the center; that is, F is the center of the circle ABC. Which was to be found.

Cor. From this it is manifest, that if in a circle a straight line bifect another at right angles, the center of the circle is in the line which bifects the other.

PRO P. II. THEO R.

IF any two points be taken in the circumference of

a circle, the straight line which joins them shall

fall within the circle.

A

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.

For if it do not, let it fall, if possible, without, as AEB; find D the center of

. I. je the circle ABC, and join AD, DB, and produce DF any straight line meeting the circumference AB, to E. then because

F DA is equal to DB, the angle DAB is A E

B equal b to the angle DBA ; and because

b. s. 1. * N. B. Whenever the expression “ straight lines from the center"

« drawn “ from the ceoter” occurs, it is .o be understood that they are drawn to the circumference.

or

c. 16. 1.

Book III. AE a fide of the triangle DAE is produced to B, the angle DED Wis greater than the angle DAE; but DAE is equal to the angle

DBE, therefore the angle DEB is greater than the angle DBE. d. 19.1.

but to the greater angle the greater side is opposited; 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 impossible. therefore tbe straight line drawn from A to B does not fall without the circle. in the same 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.

IF

equal

PROP. III. THEO R.
F a straight line drawn thro’ the center of a circle,

bisect a straight line in it which does not pass thro' the center, it shall cut it at right angles. and if it cuts it at right angles, it shall bisect it.

Let ABC be a circle ; and let CD a straight line drawn thro' the center bisect any straight line AB, which does not pass thro’

the center, in the point F. it cuts it also at right angles. 2. 1. 3

Take . E the center 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. and the base EA is equal to the

base EB ; therefore the angle AFE is
0. 8. 1. o to the angle BFE. but when a straight

line standing upon another makes the au-
jacent angles equal to one another, each

EO
C. 10.Def.1. of them is a right angle. therefore each

of the angles AFE, BFE is a right angle;
wherefore the straight line CD drawn thro'
the center bisecting another AB that does
not pass thro’ the center, cuts the same at D
right angles.

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

The same construction being made, because EA, EB from the d. 5.1.

center 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 there are two an.

gles in one equal to two angles in the other, and the side EF which Bookl II. is opposite to one of the equal angles in each, is common to both; m therefore the other sides are equal", AF therefore is equal to FB.c. 16, L. Wherefore if a straight line, &c. Q. E. D.

PROP. IV. THE OR.

IF in a

Fin a circle two straight lines cut one another which

do not both pass thro' the center, they do not bifect 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 thro'
the center. AC, BD do not bifect one another.

For, if it is possible, let AE be equal to EC, and BE to ED. if
one of the lines pass thro’ the center, it is plain that it cannot be
bifected by the other which does not
pass thro' the center. but if neither of
them pass thro’the center, take ' F the

2. r. S. center of the circle, and join EF. and

F
because FE a straight line thro' the
center, bisects another AC which does A
not pass thro' the center, it shall cut it

BS
at right bangles; wherefore FEA is a
right angle. again, because the straight
line FE bifects the straight line BD which does not pass thro' the
center, it shall cut it at right 6 angles; wherefore FEB is a right
angle. and FEA was shewn to be a right angle; therefore FEA is
equal to the angle FEB, the less to the greater, which is impos-
fible. therefore AC, BD do not bifect one another. Wherefore
If in a circle, &c. Q. E. D.

to

b. 3. 3.

1

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

the same center.

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

G

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Book III. For, if it be possible, let E be their center; join EC, and draw

any straight line EFG meeting
them in F and G. and because E

c
is the center of the circle ABC,
CE is equal to EF. again, because
E is the center of the circle CDG,
CE is equal tu EG. but CE was

E
shewn to be equal to EF; there-
fore EF is equal to EG, the less
to the greater, which is impossible.
therefore E is not the center of

B
the circles ABC, CDG. Wherefore if two circles, &c. Q. E. D.

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IF

F two circles touch one another internally, they shall

not have the same center.

Let the two circles ABC, CDE touch one another internally in the point C. they have not the same center.

For if they can, let it be F; join FC, and draw any straight
line FEB meeting them in E and B.

C
and because F is the center of the
circle ABC, CF is equal to FB. also
because F is the center of the circle
CDE, CF is equal to FE. and CF

B
was shewn equal to FB ; therefore Al

F

E
FE is equal to FB, the less to the
greater, which is impossible. where-

D
fore F is not the center of the cir-
cles ABC, CDE. Therefore if two circles, &c. Q. E. D.

Book III.

PROP. VII. THEOR.
IF any point be taken in the diameter of a circle

, which is not the center, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the center is, and the other part

of that diameter is the leaft; and of any others, that which is nearer to the line which pafses thro' the center is always greater than one more remote. 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

any point F be taken which is not the center. let the center be E; of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA is the greatest, and FD the other part of the diameter AD is the least; and of the others, FB is greater than FC, and FC than FG.

Join BE, CE, GE; and because two sides of a triangle are greater than the third, BE, EF are greater than BF; but AE is a. 26. equal to EB, therefore AE, EF, that

A is AF, is greater than BF. again, be B cause BE is equal to CE, and FE

С common to the triangles BEF, CEF; the two sides BE, EF are equal to the

E two CE, EF; but the angle BEF is greater than the angle CEF, therefore the base BF is greater than the base FC. for the same 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 greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF.

Also there can be drawn only two equal straight lines from the point I to the circumference, one upon each side of the fortest line

6. 24.

1) H

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