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For if it be not, let, if poffible, G be the center, and join GA, Book III. GD, GB. then because D A is equal to D B, and DG common

to the two triangles ADG, BDG, the two fides AD, DG are equal to the two BD, DG, each to each; and the

A

F

C

c. 8. 1.

D

B

E

bafe GA is equal to the bafe GB, because they are drawn from the center G. therefore the angle ADG is equal 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 4. therefore the angle GDB is a right angle. but FDB is like- d. 10.Def.1. wife 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 center of the circle ABC. in the fame 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.

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

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PROP. II. THEOR.

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

For if it do not, let it fall, if poffible, without, as AEB; find D the center of the circle ABC, and join AD, DB, and produce DF any straight 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

C

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"drawn

N. B. Whenever the expreffion" ftraight lines from the center" or

" from the center" occurs, it is to be understood that they are drawn to the cir

cumference.

AE

c

Book III. AE a fide of the triangle DAE is produced to B, the angle DEB is greater than the angle DAE; but DAE is equal to the angle DBE, c. 16. 1. therefore the angle DEB is greater than the angle DBE. but to the a. 19. 1. greater angle the greater fide is oppofited; 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 impoffible. therefore the straight 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.

b. 8. I.

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Fa ftraght line drawn thro' the center of a circle, bifect a straight line in it which does not pass thro' the center, it fall 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 ftraight line drawn thro' the center bifect any straight line AB, which does not pass thro' the center, in the point F. it cuts it alfo at right angles.

a

C

Take E the center of the circle, and join EA, EB. then be cause 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 base EB; therefore the angle AFE is equal to the angle BFE. but when a ftraight line standing upon another makes the adjacent angles equal to one another, each of them c.10.Def.1.is a right angle. therefore each of the

c

E

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d. 5. .

right angles.

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

The fame conftruction being made, because EA, EB from the center are cqual 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

gles in one equal to two angles in the other, and the fide EF which Book III. is opposite 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 line, &c. Q. E. D.

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PROP. IV. THEOR.

[F in a circle two straight lines cut one another which do not both pafs thro' the center, they do not bifect 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 país thro' the center. 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 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 pafs thro' the center, take F the center of the circle, and join EF. and because FE a straight line thro' the center, bifects another AC which does not pass thro' the center, it fhall cut it at right angles; wherefore FEA is a right angle. again, because the straight

b

A

B

c. 26. r.

2. 1.

F

D

E

line FE bifects the straight line BD which does not pass thro' the center, it shall cut it at right angles; 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.

PROP. V. THEOR.

F two circles cut one another, they fhall not have the fame center.

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

b.3.3

For,

Book III.

For, if it be poffible, let E be their center; join EC, and draw any ftraight line EFG meeting

them in F and G. and because E
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 to EG. but CE was
fhewn to be equal to EF; there-
fore EF is equal to EG, the lefs
to the greater, which is impoffi-
ble. therefore E is not the cen-

G

A

D

F

E

B

ter of the circles ABC, CDG. Wherefore if two circles, &c. Q. E. Di

PROP. VI. THEOR.

Ftwo circles touch one another internally, they shall not

IF

have the fame center.

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

For if they can, let it be F; join FC, and draw any straight line FEB mecting them in E and B. and because F is the center of the circle ABC, CF is equal to FB. alfo because F is the center of the circle CDE, CF is equal to FE. and CF was fhewn cqual to FB; therefore A FE is equal to FB, the lefs to the

F

/E

greater, which is impoffible. where-
fore F is not the center of the circles
ABC, CDE.

D

Therefore if two circles, &c. Q. E. D.

B

PROP

PROP. VII. THEOR.

Fany 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 paffes thro' the center is always greater than one more remote. and from the fame point there can be drawn only two straight lines that are equal to one another, one upon each fide of the fhorteft line.

Let ABCD be a circle, and AD its diameter, in which let 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 leaft; and of the others, FB is greater than FC, and FC than FG.

Book III.

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

BA

C

E

S. 24. 1.

K

G

D

H

is AF, is greater than EF. again, be
caufe BE is equal to CE, and FE
common to the triangles 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 reafon,
CF is greater than GF. again, be-
caufe 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. there-
fore FA is the greatest, and FD the leaft 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 ftraight lines from the point F to the circumference, one upon each fide of the fhorteft line

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