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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 fides AD, DG are e

C qual to the two BD, DG, each to each ; and the base GA is equal to the base GB, because they are drawn from the centre G*: Therefore the

F G angle ADG is equal to the angle

« 8.1. GÖB: But when a straight line standing upon another straight line makes A

B

D the adjacent angles equal to one another, each of the angles is a right an.

E gled: Therefore the angle GDB is a

dro, def. 1. 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 impoffible: Therefore G is not the centre of the cir. cle ABC : In the same manner it can be shewn, that no other point but F is the centre; that is, Fis the centre of the cirde ABC: Which was to be found.

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 bisects the other.

PRO P. II. THE O R.
F any two points be taken in the circumference of a

circle, the straight line which joins them shall fall within the circle.

Let ABC be a circle, and A, B any two points in the cir• cumference; tbe straight line drawn

С 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 cen.

a I. 3. tre of the circle ABC, and join AD,

D DB, and produce DF, any straight line meeting the circumference AB, to E: Then because DA is equal to DB, the F angle DAB is equal-to the angle DBA;

bs. Sa and because AE, a fide of the triangle A E B E 3

DAE, • 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 cira cumference.

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

angle DAÉ; but DAE is equal to the angle DBE; therefore ( 16. I.

the angle DEB is greater than the angle DBE: But to the greatd 19. I er angle the greater fide 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 imposible: 'Therefore the sraight 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 with in it. Wherefore, if any two points, Sc Q. E. D.

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IF. a straight line drawn through the centre of a circle

bifect a Itraight line in it which does not pass through the centre, 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 through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It cuts it also at right

angles. a 1. 3.

Take . E the centre of the circle, and join EA, EB: Then, because AF is equal to FB, and FE common to the two tri. Engles AFE, BFE, there are two fides in the one equal to two sides in the other, and the base EA is

с
equal to the base EB ; therefore the
b 8. I.

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 a right

E c 19. def. č. € angle : Therefore each of the angles

AFE, BFE iş a right angle; wherefore
the straight line CD, drawn through the

F
centre bisecting another AB that does

А
not pass through the centre, çuts the samo P
at right angles.

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

The same construction bciog made, because EA, EB from 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: Therciore, in the iwo triangles EAF, EBF, Hi there are two angles in one equal to two angles in the other, Book III. and the Gide EF, which is opposite to one of the equal angles

there

in each, is common to both; therefore the other sides are eA qual; AF therefore is equal to FB. Wherefore, if a straight c 26. 1.

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 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 passthrough the centre: But, if neither of them pass through the centre, take F the centre of the circle, and

a I. 3• join EF: And because Fe, a straight

F D line through the centre, bisects another A AC which does not pass through the centre, it shall cut it at right angles; B E C +3° 3. wherefore FEA is a right angle : Again, because the straight line FE bilects the straight line BD which does not pass through the centre, it shall cut it at right 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 im. possible : Therefore AC, BD do not bisect one another. Where-, fore, if in a circle, &c. Q. E. D.

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

same centre.

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

E 4

For,

Book III.

G

For, if it be possible, let E be their centre : Join EC, and draw any straight line EFG meet

С.
ing them in F and G: And because
E is the centre of the circle ABC,
CE is equal to EF : Again, be:
cause E is the centre of the circle

А
CDG, CE is equal to EG: But D

F
CE was fhewn to be equal to EF;

E
therefore EF is equal to EG, the
Jess to the greater, which is impos-
fible: Therefore £ is not the cen-
tre of the circles ABC, CDG.
Wherefore, if two circles, &c. Q. E. D.

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IF two circles touch one another internally, they shall

not 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 can, let it be F; join FC and draw any straight line FEB meeting them in E and B;

0
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

B
CF was fhewn equal to FB ; there.

E

T
fore FE is equal to FB, the less to A
the greater, which is impossible:
Wherefore F is not the centre of the

D
circles ABC, CDE. Therefore, if
Wo circles, &c. Q. E. D.

PROP

Blok III.

PRO P. VII." THE O R.

any point be taken in the diameter of a circle, which

is not the centre, of all the straight 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 diametet is the least ; and, of any others, that which is nearer to the line which passes through the centre 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 let any point F be taken which is not the centre : Let the centre 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. 20. I. is equal to EB ; therefore AE, EF, that is AF, is greater than BF: A. B

A gain, because BE is equal to CE, and FE common to the triangles C BEF, CEF, the two fides BE, EF are equal to the two CE, EF; but

E the angle BEF is greater than the angle CEF; therefore the base BF is greater than the base FC: For the

b 24. I. lame reason, CF is greater than GF:

K Again, because GF, FE are greater

H * 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 Atraight lines From F to the circumference, and BF is greater chan CF, and CF than GF.

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

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