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Book Ill.

PROP. IX. THEOR.

a

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.

For, if not, let E be the centre,
join DE and produce it to the cir-
cumference in F, G: then FG is
a diameter of the circle ABC: And
because in FG, the diameter of the F

DE
circle ABC, there is taken the point
D, which is not the centre, DG

shall be the greatest line from it to • 2. 3. the circumference, and DC greater a

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

PROP. X. THEOR,

BD

ONE circumference of a circle cannot cut another
in more than two points.
If it be possible, let the circum-

А
ference FAB cut the circumfer-
ence DEF in more than two B.

H
points, viz. in B, G, F; take the
centre K of the circle ABC, and

EK!
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,

a 9. 3.

KG, KF, the point K isa the centre of the circle DEF: Boox III. 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 impossibleb. Therefore one circumference of a 15.3. circle cannot cut another in more than two points. Q.E.D.

PROP. XI. THEOR.

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 ABC, ADE touch each other internally in the point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE: 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 possible, as FGDH, and join AF, AG: And because AG, GF are greater a than FA, that is, than FH, for FA is equal to FH, both being from the

E same centre; take away

the common

B 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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PROP. XII. THEOR.

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 ABC, ADE, touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of ADE: 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,

Book III

. and join FA, AG: And because F is the centre of the cit. cle ABC, AF is equal

E
to FC: Also, hecause

B
G is the centre of the
circle ADE, AG is

A
equal to GD: There-
fore FA, AG are equal
to FC, DG; where-
fore the whole FG is

greater than FA, AG: * 20.1. But it is also lessa; 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.

PROP. XIII. THEOR. See N. ONE circle cannot touch another in more points

than one, whether it touches it on the inside or outside.

For, if it be possibile, let the circle EBF touch the circle

ABC in more points than one, and first on the inside, in • 10. 11. 1. the points B, D;

join BD, and drawa GH bisecting BD at right angles : Therefore because the points B, D, are in

H
A

A
E

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D

С

с

F

G the circumference of each of the circles, the straight line b 2. 3. • Cor. 1. 3.

BD falls within each b of them : and their centres are in

the straight line GH which bisects BD at right angles : d 11. 3. Therefore GH passes through the point of contact d; but it

does not pass through it, because the points B, D are without the straight line GH, 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 Book III. in more than one point; For, if it be possible, let the circle ACK touch the circle ABC in the points A, C, and join AC: Therefore,

K because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them shall fall within the circle ACK: And

12. 3. the circle ACK is without the circle ABC; and therefore the straight line AC is without this last circle; but because the points A, C are in the circumference of the circle ABC, the straight line AC must be within the B same circle, which is absurd : Therefore one circle cannot touch another on the outside in more than one point. And it has been shown, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q.E.D.

a

PROP. XIV. THEOR. 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 AB, CD, in the circle ABDC, be equal to one another, they are equally distant from the centre.

Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD: Then, because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects ait: Wherefore AF is A equal to FB, and AB double of AF. For the same reason CD is double of CG: And AB is equal to CD; therefore AF is equal to CG: And because AE is equal to EC, the square of AE is equal to the square of EC: But

:

BT the squares of AF, FE, are equalb to

b 17. 1. the square of AE, because the angle AFE is a right angle; and for the like reason, the squares of EG, GC are equal to the square of EC: Therefore the squares of AF, FE, are

a 3. 3.

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Book III. equal to the squares of CG, GE, of which the square of AF

is equal to the square of CG; because AF is equal to CG;
therefore the remaining square of FE is equal to the re-
maining square of EG, and the straight line EF is therefore
equal to EG: But straight lines in a circle are said to be

a
equally distant from the centre, when the perpendiculars
€ 4 Def. 3. drawn to them from the centre are equalc: Therefore AB,

CD must be equally distant from the centre,

Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG; AB is equal to CD: For, the same construction being made, it

may, as before, be demonstrated, that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG;

wherefore AB is equal to CD. Therefore equal straight lines, &c. Q.E.D.

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See n. The diameter is the greatest straight line in a cir

cle; and, of all others, that which is nearer to the
centre is always greater than one more remote: and
the

greater is nearer to the centre than the less.
Let ABCD be a circle, of which

А В
the diameter is AD, and the centre

E
E; and let BC be nearer to the cen-
tre than FG; AD is greater than any
straight line BC, which is not a dia-

H

Kmeter, and BC greater than FG.

E
From the centre draw EH, EK
perpendiculars to BC, FG, and join

G
EB, EC, EF; and because AE is

С
equal to EB, and ED to EC, AD is
* 20. 1. equal to EB, EC; but EB, EC are greatera than BC:

wherefore also AD is greater than BC.

And, because BC is nearer to the centre than FG, EH

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