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PROPOSITION III.—THEOREM.

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 a b, which does not pass through the centre, in the point f; it cuts it also at right angles.

Take (i. 3) e the centre of the circle, and join e a, e b. 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 e a is equal to the base e b; therefore the angle a fe is equal (i. 8) 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 angle (i. def. 10): therefore each of the angles afe, bfe, is a right angle; wherefore the straight line cd, drawn through the centre, bisecting another ab that does not pass through the centre, cuts the same at right angles.

d

But let c d cut ab at right angles; c d also bisects it; that is, af is equal to fb.

The same construction being made, because ea, e b, from the centre are equal to one another, the angle e a f is equal (i. 5) to the angle ebf: and the right angle a fe is equal to the right angle bfe: therefore, in the two triangles e af, e bf, there are two angles in 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 (i. 26); af therefore is equal to fb. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION IV.-THEOREM.

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 a b c d be a circle, and a c, b d two straight lines in it which cut one another in the point e, and do not both pass through the centre: a c, b d, do not bisect one another.

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d

For, if it be possible, let a e be equal to e c, 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 (i. 3) f the centre of the circle, and join ef; and because fe, a straight line through the centre, bisects another a c which does not pass through the centre, it shall cut it at right (iii. 3) 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 shall cut it at right angles (iii. 3): 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 impossible therefore a c, bd do not bisect one another. Wherefore, if in a circle, &c. Q. E. D.

PROPOSITION V.—THEOREM.

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

LET the two circles a bc, cdg cut one another in the points b, c; they

have not the same centre.

For, if it be possible, let e be their centre; join ec, and draw any straight line efg meeting them in fand g; and because e is the centre of the circle a b c ce is equal to ef. Again, because e is the centre of the circle cdg, ce is equal to eg: but ce was shewn to be equal to ef: therefore e f 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.

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PROPOSITION VI.-THEOREM.

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 is equal to fe: and cf was shewn to be equal to fb; therefore fe is a equal to fb, the less to the greater, which is impossible wherefore f is not the centre of the circles abc, cde. Therefore, if two circles, &c. Q. E. D.

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PROPOSITION VII.-THEOREM.

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 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 (i. 20) than the third, be, ef, are greater than bf; but ae is equal to

a

eb; therefore a e, ef, that is af, is 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 greater (i. 24) than the base fc: for the same reason cf is greater than gf. Again, because gf, fe are greater (i. 20) 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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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 straight line ef, make (i. 23) 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; therefore the base fg is equal (i. 4) 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, 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.

PROPOSITION VIII.-THEOREM.

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 apon 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 upon the concave part of the circumference ae fc, 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 h1kg, the least is dg between the point d and the diameter a g; and the nearer to it is always less than the more remote, viz. dk than dl, and d1 than dh.

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gb

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Take (i. 3) m the centre of the circle a bc, and join me, mf, mc, mk, ml, mh. And because a m is equal to me, add md to each, therefore ad is equal to em, md; but em, md are greater (i. 20) than ed; therefore also ad is greater than ed. Again, because me is equal to mf, and md common to the triangles em d, 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 (i. 24). In like manner it may be shewn that fd is greater than cd. Therefore da is the greatest, and de greater than df, and df than dc. And because mk, kd are greater (i. 20) than md, and mk is equal to mg, the remainder k d is greater (4 ax.) than the remainder gd, that is, gd is less than k d. And because mk, dk are drawn to the point k within the triangle mld from m, d, the extremities of its side md; mk, kd, are less (i. 21) 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 shewn that dl is less than dh: therefore dg is the least, and dk less than dl, and d1 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 dm b 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 km d is equal to the angle bmd; therefore the base dk is equal (i. 4) to the 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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If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle.

LET the point d be taken within the circle abc, from which to the circumference there fall more than two equal straight lines, viz. da, db, dc, the point d is the centre of the circle.

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de

For, if not, let e be the centre, join de and produce it to the circumference in fg then fg is a diameter of the circle abc. And because in fg, the diameter of the circle a bc, there is taken the point d, which is not the centre, dg shall be the greatest line from it to the circumference, and dc greater (iii. 7) than db, and db than da. But they are likewise equal, which is impossible; therefore e is not the centre of the circle abc. In like manner it may be demonstrated that no other point but d is the centre; d therefore is

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the centre. Wherefore, if a point be taken, &c. Q. E. D,

PROPOSITION X.-THEOREM.

One circumference of a circle cannot cut another in more than two points. If it be possible, let the circumference abc cut the circumference def

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in more than two points, viz. in b, g, f; take the centre k of the circle abc, and join kb, kg, kf. And because within the circle def there is taken the point k, from which to the circumference def fall more than two equal straight lines kb, kg, kf, the point k is the centre of the circle def (iii.9). But k is also the centre of the circle abc; therefore the same point is the centre of two circles that cut one another, which is impossible (iii. 5). Therefore one circumference of a circle cannot cut another in more than two points. Q. E. D.

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