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there are two angles in one equal to two angles in the other, Book III. and the fide EF, which is oppofite to one of the equal angles in each, is common to both; therefore the other fides are equale; AF therefore is equal to FB. Wherefore, if a straight e 26. 1. line, &c. Q. E. D.

PROP. IV. THEO R.

IF in a circle two straight lines cut one another which do not both pafs through the centre, they do not bifect each 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 pass through the centre: AC, BD do not bisect one another.

For, if it is poffible, let AE be equal to EC, and BE to ED: If one of the lines pafs through the centre, it is plain that it cannot be bifected by the other which does not pass through the centre: But if neither of them pass through the centre, take a F the centre of the circle, and join EF: and because FE, a ftraight line through the centre, bi- A fects another AC which does not pafs through the centre, it fhall cut it at right bangles; wherefore FEA is a right angle: Again, because the

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ftraight line FE bifects the straight line BD which does not pass through the centre, it fhall cut it at right bangles; wherefore FEB is a right angle: And FEA was shown 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 bisect 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 centre.

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

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Book III.

For, if it be poffible, let E be their centre: Join EC, and draw any ftraight line EFG meeting them in F and G: and because E is the centre of the circle ABC, CE is equal to EF: Again, becaufe E is the centre of the circle A CDG, CE is equal to EG: But CE was shown to be equal to EF; therefore EF is equal to EG, the lefs to the greater, which is impoffible: Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D.

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PROP. VI. THEOR,

two circles touch one another internally, they fhall not have the fame centre.

Let the two circles ABC, CDE, touch one another internally in the point C: They have not the fame centre.

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

And because F is the centre of the
circle ABC, CF is equal to FB:
Alfo, because F is the centre of the
circle CDE, CF is equal to FE:
And CF was fhewn equal to FB;
therefore FE is equal to FB, the lefs
to the greater, which is impoffible;
Wherefore F is not the centre of
the circles ABC, CDE. Therefore,
if two circles, &c. Q. E. D.

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PROP. VII. THEOR.

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 greateft is that in which the centre is, and the other part of that diameter is the leaft; and, of any others, that which is nearer to the line which paffes through the céntre is always greater than one more remote: And from the fame point there can be drawn only two ftraight lines that are equal to one another, one upon each fide 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 ftraight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA is the greateft, and FD, the other part of the diameter BD, is the least: And of the others, FB is greater than FC, and FC than FG.

Book III.

Join BE, CE, GE; and because two fides of a triangle are greater a than the third, BE, EF are greater than BF; but AE a 20: 1. is equal to EB; therefore AE,EF, that is AF, is greater than BF: Again, 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 the angle BEF is greater than the angle CEF; therefore the bafe BF is greater b than the base FC: For the fame reason, CF is greater than GF: Again, because GF, FE are greater a than EG, and EG is equal

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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 greateft, and FD the least of all the ftraight 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

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4. I.

Book II. fhorteft line FD: At the point E in the straight line EF, make the angle FEH equal to the angle GEF, and join FH: Then becaufe GE is equal to EH, and EF common to the two triangles GEF, HEF; the two fides GE, EF are equal to the two HE, EF; and the angle GEF is equal to the angle HEF; therefore the bafe FG is equal d to the bafe FH: But, befides 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 be caufe FK is equal to FG, and FG to FH, FK is equal to FH; that is, a line nearer to that which paffes through the centre, is equal to one which is more remote; which is impoffible. Therefore, if any point be taken, &c. Q. E. D.

PROP. VIII. THEOR.

IF any point be taken without a circle, and ftraight

lines be drawn from it to the circumference, whereof one paffes through the centre of those which fall upon the concave circumference, the greatest is that which paffes 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 leaft is that between the point without the circle, and the diameter; and of the reft, that which is nearer to the leaft is always lefs than the more remote :. And only two equal straight lines can be drawn from the point unto the circumference, one upon each fide of the leaft.

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 paffes through the centre. . Of those which fall upon the concave part of the circumference AEFC, the greatest is AD which paffes 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 leaft is DG between the

point

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point D'and the diameter AG; and the nearer to it is always Book III. lefs than the more remote, viz. DK than DL, and DL than

DH.

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d 4. Ax?

Take a M the centre of the circlc ABC, and join ME, MF, a 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 e greater b than ED; therefore alfo AD is greater than ED: b 20. 1. Again, because ME is equal to MF, and MD common to the triangles 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 c than the base FD: In like manner it may be shewn that FD is greater than CD: Therefore DA is the greatest; and DE greater than DF, and DFthan DC: And because MK, KD are greater b than MD, and MK is 'equal to MG, the remainder KD is greater d than the remainder GD, that is GD is lefs than KD: And because MK, DK are drawn to the point K within the triangle MLD from M. D, the extremities of its fide MD, MK, KD are lefse than ML,LD, whereof MK is equal to ML; therefore the remainder DK is lefs than the remainder DL: In like manner it may be fhewn, that DL is lefs than DH: Therefore DG is the leaft, and DK lefs than DL,and DL than DH: Alfo there can be drawn only two equal straight lines from the point D to the circumference, one upon each fide of the leaft: 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 fides 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 f to the base DB: But, befides DB, f 4. r. there can be no ftraight 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 alfo to DB; therefore DB is equal to DN, that is, the nearer to the leaft equal to the more remote, which is impoffible. If therefore, any point, &c. Q. E. D.

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