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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 a d 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 a d, 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 a e is equal to

e b; therefore a e, ef, that is a f, 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 g f. fe are greater (i. 20) than and

is equal to ed, gf, fe are greater than ed; take k

away the common part f e, and the remainder g.f d

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

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 fe h equal to the angle gef, and join fh: then, because ge is equal to eh, and e f common to the two triangles gef, hef; the two sides ge, ef are equal to the two he, ef; and the angle g ef 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, 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.

eg

eg

8

than gf.

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 tapon each side of the least.

meter a g;

m

n

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 a efc, the greatest is a d 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 blkg, the least is dg between the point d and the dia

and the nearer to it is always less than the more remote, viz. dk than dl, and dl than dh.

Take (i. 3) m the centre of the circle a b c, and join me, mf, mc, mk, ml, mh. And because am is equal to me, add m d to each, therefore ad is equal to em md; but em, m d are greater (i. 20) than ed; therefore also a d is greater than ed. Again, because me is equal to

d 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 than the base fd (i. 24). In like manner it may be shewn that fd is greater

k/

gb than cd. Therefore da is the greatest, and de greater than df, and df than dc. And because mk, k d 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, k d, are less (i. 21) than ml, ld, whereof mk is equal to ml; therefore the remainder dk is less than the remainder dl. In

a like manner it may be shewn 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 d b. And because mk is equal to mb, and md common to the triangles k md, bmd, the two sides km, m d are equal to the two b m, md, and the angle k md is equal to the angle b md; therefore the base dk is equal (i. 4) to the base db. But besides d b 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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m

PROPOSITION IX.-THEOREM.

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.

e

LET the point d be taken within the circle a b c, 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. For, if not, let e be the centre, join de and produce it to the circum

ference in f, g: then fg is a diameter of the circle a b c. And because in fg, the diameter of the circle a b c, there is taken the point d,

which is not the centre, dg shall be the greatest f

8 line from it to the circumference, and dc greater

(iii. 7) than d b, and db than da. But they are likewise equal, which is impossible; therefore e is not the centre of the circle a b c. In

like manner it may be demonstrated that no b

other point but d is the centre; d therefore is the centre. Wherefore, if a point be taken, &c. Q. E. D.

de

C

PROPOSITION X-THEOREM.

b

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

in more than two points, viz. in b, g, f; take

the centre k of the circle a b c, and join k b, d

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 k b, kg, kf, the point k is the centre of the circle d e f (iii. 9). But k is also the

centre of the circle abc; therefore the same f

point is the centre of two circles that cut one another, which is impossible (iii. 5). There

5 fore one circumference of a circle cannot cut another in more than two points. Q. E. D.

09

PROPOSITION XI.-THEOREM. If two circles touch each other internally, the straight line which joins their

centres being produced shall pass through the point of contact. LET the two circles a b c, ade touch each other internally in the point a, and let f be the centre of the circle a b c, and

8
the centre of the circle a de.

a
The straight line which joins the centres
f, g, being produced, passes through the
point a

h For, if not, let it fall otherwise, if posşible, as fgdh, and join af, ag. And

a because ag, gf are greater (i. 20) than fa, that is, than fh, for fa is equal to fh, both being from the same centre; take away the common part fg; therefore the remainder ag is greater than the remainder gh. But ag is equal to gd;

. therefore gd is greater than gh, the less than the greater, which is impossible. Therefore the straight line which joins the points f, g cannot fall otherwise than upon the point a, that is, it must pass through it. Therefore, , if two circles, &c. Q. E. D.

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PROPOSITION XII.—THEOREM. If two circles touch each other externally, the straight line which joins their

centres shall pass through the point of contact. LET the two circles a b c, a de touch each other externally in the point a; and let f be the centre of the circle a b c, and g the centre of a de. The straight line which joins the points f, g shall pass through the point of contact a.

For, if not, let it pass otherwise, if possible, as fcdg, and join

b And because f is the centre of the circle a b c, af is equal to fc. Also, because g is the centre of the circle a de, ag

f is equal to gd. Therefore fà,

8 ag are equal to fc, dg; wherefore the whole fg is greater than fa, ag: But it is also less (i. 20); which is impossible. Therefore the straight line which joins the points f, g, shall not pass otherwise than through the point of contact a, that is, it must pass through it. Therefore, if two circles, &c. Q. E. D.

fa, ag.

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PROPOSITION XIII.-THEOREM.

One circle cannot touch another in more points than one, whether it touches

it on the inside or outside. For, if it be possible, let the circle e bf touch the circle a bc in more points than one, and first on the inside, in the points b, d; join bd, and draw (i. 10, 11) g h bisecting bd at right angles. Therefore, because

h e

a

b

la

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gh,

8 the points b, d are in the circumference of each of the circles, the straight line bd falls within each of them (iii. 2): and their centres are in the straight line gh which bisects b d at right angles (iii. 1. Cor.). Therefore gh passes through the point of contact (iii. 11); but it does not pass through it, because the points b, d are without the straight line which is absurd. Therefore one circle cannot touch another on the inside in more points than one.

Nor can two circles touch one another on the outside in more than one point; for, if it be possible, let the circle ack touch the circle a b c

in the points a, C, and join a c. Therefore, because k the two points a, c, are in the circumference of the

circle ack, the straight line a c which joins them shall fall within the circle ack (iii. 2). And the circle a ck is without the circle a bc; and therefore the straight line ac is without this last circle ; but because the points a, c are in the circumference of the circle a b c, the straight line ac must be within (iii. 2) the same circle, which is absurd. Therefore one circle cannot touch another on the outside in

more than one point. And it has been shewn that 'b they cannot touch on the inside in more points than

Therefore, one circle, &c. Q. E. D.

a

a

one.

PROPOSITION XIV.—THEOREM. Equal straight lines in a circle are equally distant from the centre, and those

which are equally distant from the centre are equal to one another. LET the straight lines a b, cd, in the circle abdc, be equal to one another, they are equally distant from the centre.

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