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b

b 8. 1.

one equal to two sides in the other; but the base EA is Book III. equal to the base EB; therefore the angle AFE is equal to the angle BFE. And when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right e angle: Therefore each of the angles

c

E

c 7. def. 1

AFE, BFE is a right angle; where

fore the straight line CD, drawn A

B

F

through the centre, bisecting AB,

which does not pass through the centre, cuts AB at right angles.

D

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

The same construction being made, because the radii EA, EB are equal to one another, the angle EAF is equal to the angle EBF; and the right angle AFE is d 5. 1. equal to the right angle BFE: Therefore, in the two triangles EAF, EBF, there are two angles in the one equal to 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 equal; AF there- e 26. 1. fore is equal to FB. Wherefore, if a straight line, &c. Q. E. D.

PROP. IV. THEOR.

If in a circle two straight lines cut one another, in a point which is not the centre, they cannot bisect each other.

Let ABCD be a circle, and AC, BD two straight lines in it, which cut one another in a point E, which is not the centre: AC, BD do not bisect one another. For, if 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 thro' A 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 straight line through the cen

F

a 1. 3.

b. 3. 3.

b

b

Book III. tre, bisects another AC, which does not pass through the centre, it must cut it at right angles; wherefore FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, it must cut it at right 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.

PROP. V. THEOR.

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If two circles cut one another, they cannot have the same centre.

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

C

For, if it be possible, let E be their centre; join EC, and draw any straight line EFG meeting the circles in F and G; and because E is the centre of the circle ABC, CE is equal to EF: Again, because E is the cen- A tre of the circle 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

G

F

E

B

greater, which is impossible: therefore E is not the centre of the circles ABC, CDG. Wherefore, if two eircles, &c. Q. E. D.

PROP. VI. THEOR.

If two circles touch one another internally, they cannot 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 Book III. straight line FEB meeting the circles 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 is equal to FE: but CF was shewn to be equal A to FB; therefore FE is equal to FB, the less to the greater, which is impossible; wherefore F is not

F

E

B

D

the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D.

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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 passing through the centre is always greater than one more remote from it: 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 tri

a

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

EF, that is, AF is greater than BF: again, because BE

A

Book III 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

a

B

E

b 24. 1. BF is greater than the base FC; for the 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.

G

H

D

Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD: at the point E in the c. 23. 1. straight 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 is equal to the angle HEF; d 4. 1. therefore the base FG is equal to the base FH: but besides FH, no 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 more remote, which is impossible. Therefore, if any point be taken, &c. Q. E. D.

d

Book III.

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 into the circumference, one upon each side of the least.

centre.

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 Of those which fall upon the concave part of the circumference AEFC, the greatest is AD which passes through the centre; and the line nearer to AD 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; and the nearer to it is always less than the more remote, viz. DK than DL, and DL than DH.

Take a M the centre of the circle ABC, and join ME, a 1. 3. MF, MC, MK, ML, MH: And because AM is equal

b

to ME, if MD be added to each, AD is equal to EM and MD; but EM and MD are greater than ED; b 20. 1. therefore also AD is greater than ED. Again, because ME is equal to MF, and MD common to the triangles

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