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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 botlı; therefore the other sides are equale; AF therefore is equal é 26. 1. to FB. Wherefore, if a straight line, &c. Q. E. D.

PROP. IV. THEOR.

Ir 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 ÉC, 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, take a F

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

B through the centre, it shall cut it at right 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 right b angles : wherefore FEB is a right angle: And FEĂ 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.

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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

G
ABC, CE is equal to EF : Again, A
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 FEBmeeting 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, CE is equal to FE: And CF was

A

E shown to be equal to FB; therefore FE is equal to FB, the less to the greater, which is impossible : Wherefore F is not the centre of the circles ABC, CDE. There fore, if two circles, &c:. Q. E. D.

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Book III.

PROP. VII. THEOR.

If 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 FR, 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

a are greater a than the third, BE, EF, are greater than BF;- 20. 1. but AE is equal to EB; therefore AE, EF, that is AF, is

B greater than BF: Again, because 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 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.

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a

فونز A

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

of the shortest line FD: At the point E in the straight < 23. 1. line EF, make the angle FEH equal to the angle GEF,

and join FH: Then because GE is equal to EH, and EF common to the two triangles GEF, HEF; the two sides

GE, EF are equal to the two HE, EF; and the angle GEF 4. 1. is equal to the angle HEF; therefore the base FG is equald

to the base FH: But, besides FH, no other straight line can be drawn from F to the circumference equal to FG: For, if there can, let it be FK; and because FK is equal to FG, and FG to FH, FK is equal to FH; that is, a line nearer to that which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if any point be taken, &c. Q.E.D.

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PROP. VIII. THEOR.

If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre ; and of the rest, that which is nearer to that through the centre is always greater than the more remote: But of those which fall upon the convex circumference, , the least is that between the point without the circle and the diameter; and of the rest, that which is nearer to the least is always less than the more remote: And only two equal straight lines can be drawn from the point unto the circumference, one upon each side of the least.

Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC, be drawn to the circumference, whereof DA passes through the centre. Of those which fall

the

concave part of the circunference AEFC, the greatest is AD which passes through the centre; and the nearer to it is always greater than the more remote, viz. DE than DF, and DF 'than DC: But of those which fall upon the convex circumference HLKG, the least is DG between the point D and the diameter AG;

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and the nearer to it is always less than the more remote, Book III. viz. DK than DL, and DL than DH.

Take a M the centre of the circle ABC, and join ME, MF, *1.3. MC, MK, ML, MH: And because AM is equal to ME, add MD to each, therefore AD is equal to EM, MD; but EM, MD are greaterb than ED; therefore also AD is ! 20. 1. greater than ED. Again, because ME is equal to MF, and MD common to the triangles

D EMD, FMD; EM, MD, are equal to FM, MD: but the angle EMD is greater than the angle FMD; therefore the base ED is greater than the base FD. In like manner it may be shown

GB that FD is greater than CD:

IN

Н.
Therefore DA is the greatest;
and DE greater than DF, and
DF than DC: And because MK,
KD are greater than MD, and .
MK is equal to MG, the remain-C

AM der KD is greater than the re

d4 Ax. mainder GD, that is GD is less than KD: And because MK,

F DK, are drawn to the point K

E within the triangle MLD, from M, D, the extremities of its side MD, MK, KD, are lesse 21. 1. than ML, LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL: In like manner it may be shown, that DL is less than DH: Therefore DG is the least, and DK less than DL, and DL than DH. Also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least. At the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB : And because MK is equal to MB, and MD common to the triangles KMD, BMD, the two sides KM, MD are equal to the two BM, MD; and the angle KMD is equal to the angle BMD; therefore the base DK is equal to the'r 4. 1. base DB: But, besides DB, there can be no straight line drawn from D to the circumference equal to DK: For, if there can, let it be DN; and because DK is equal to DN, and also to DB; therefore DB is equal to DN, that is, the nearer to the least equal to the more remote, which is impossible. If therefore any point, &c. Q.E.D.

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