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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 sides AD, DG, are equal 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 angle ADG is equal to the angle

• 8. 1. GDB : But when a straight line standing upon another straight line

AZB makes the adjacent angles equal to one another, each of the angles is a right angled : Therefore the angle GDB is a right angle: " 10 Def. 1. 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, shat 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 shall fall within the circle. .

Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from Ato B shallfall within the circle. ..

For, if it do not, let it fall, if possi. ble, without, as AEB; find a Ď the centre 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 equalh to the angle DBA; and because AE,

* 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.

5. i.

C 16.

Book III. a side 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; therefore the angle DEB is greater than

the angle DBE: But to the greater angle the greater side 19. 1. 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 the 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.

PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect a straight 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. 13. Takea E the centre 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; there-
58. 1. fore the angle AFE is equalb to the

angle BFE: But when a straight
line standing upon another makes the

adjacent angles equal to one another, < 10 Def. 1. each of them is a right angle: There

fore each of the angles AFE, BFE, is
a right angle; wherefore the straight A
line CD, drawn through the centre,
bisecting another AB that does not
pass through the centre, cuts the same at right angles.

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

The same construction being made, because EA, EB,

from the centre are equal to one another, the angle EAF. d 5. 1. is equald to the angle EBF: and the right angle AFE is

equal to the right angle BFE: Therefore, in the two tri

anglangles in the otheangles in each; AF therefore is.

angles, EAF, EBF, there are two angles in one equal to Book III. two angles in the other, and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equale; AF therefore is equal • 26. 1. to FB. Wherefore, if a 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 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, takea F

na 1. 3. the centre of the circle, and join AL. EF; and because FE, a straight line through the centre, bisects another AC which does not pass through the centre, it shall cut it at rightb angles; where. b 3. 3. fore FEA is a right angle: Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, it shall cut it at rightb 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.

E

PROP. V. THEOR. 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.

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

draw any straight line EFG meet-
ing them in F and G; and be-
cause E is the centre of the circle
ABC, CE is equal to EF: Again, All
because E is the centre of the cir-
(cle CDG, CE is equal to EG:
But CE was shown to be equal to
EF; therefore EF is equal to EG,
the less to the greater, which is
impossible : Therefore E is not
the centre of the circles ABC, CDG. Wherefore, if two
circles, &c. Q. E. D.

PROP. VI. THEOR. 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 have, 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; Also, because F is the centre of the circle CDE, CF A is equal to FE: And CF was. shown to be equal to FB; therefore FE is equal to FB, the less to s the greater, which is impossible : Wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D.

Book III.

PROP. VII. THEOR. Ir 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 diameter 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 other, 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; • 20. 1. but AE is equal to EB; therefore AE, EF, that is AF, is greater than BF: Again, be.. cause BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF; but the angle BEF is greater than the angle CEF; therefore the base BF is greaterb than the base FC: For the

» 24. 1. same reason, CF is greater than GF: Again, because GF, FE are greatera 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.

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