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c

D

GBN

Book III. EMD, FMD; EM, MD are equal to FM, MD; but the angle EMD is greater than the angle FMD; therec 24. 1. fore the base ED is greater 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 DF than DC.

d

And because MK, KD are greater than MD, and MK is equal to MG, the remain- C d 5. Ax, der KD is greater than the remainder GD, that is GD is less than KD: And because MK, DK are drawn to the point K within the

F

LK GB

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triangle MLD from M, D, the extremities of its side e 21. 1. MD; MK, KD are less 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 DL than DH.

f 4. 1.

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 DB; and because in the triangles KMD, BMD, the side KM is equal to the side BM, and MD common to both, and also the angle KMD equal to the angle BMD, the base DK is equal to the base DB. But, besides DB, no straight line can be drawn from D to the circumference, equal to DK: for, if there can, let it 'be DN: then, because DN is equal to DK, and DK equal to DB, DB is equal to DN; that is, the line nearer to DG, the least, equal to the more remote, which has been shewn to be impossible. If, therefore, any point, &c. Q. E. D.

Book III.

If

PROP. IX. THEOR.

a point be taken within a circle, from which there fall more than two equal straight lines upon the circumference, that point is the centre of the circle.

Let the point D be taken within the circle ABC, from which there fall on the circumference 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 circumference in F, G;

then FG is a diameter of the
circle ABC: And because in
FG, the diameter of the circle

ABC, there is taken the point F DE
D which is not the centre, DG
is the greatest line from it to the
circumference, and DC greater

A

B

G

a than DB, and DB than DA;31
but they are likewise equal, d
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. Wherefore, if
a point be taken, &c. Q. E. D.

a 7, 3.

PROP. X. THEOR.

One circle cannot cut another in more than two points.

If it be possible, let the circumference FAB cut the circumference DEF 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

Book III. there is taken the point K, from which more than two

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equal straight lines, viz. KB,
KG, KF, fall on the circum-
ference DEF, the point K is "
the centre of the circle DEF.;
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

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b 5. 3. impossible. Therefore one cir-
cumference of a circle cannot
cut another in more than two points,

C

Q. E. D.

PROP. XI. THEOR.

If two circles touch each other internally, the straight line which joins their centres being produced, will pass through the point of con

tact.

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, be- H
ing produced, passes through the
point A.

a

XD

F

E

B

For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG, GF a 20. 1. are greater than FA, that is, than FH, for FA is equal to FH, being radii of the same circle; take away the common part FG, and the remainder AG is greater than the remainder GH. But AG is equal to GD, therefore GD is greater than GH; and it is also less, which is impossible. Therefore the straight line which joins the points F and G cannot fall otherwise than on the point A; that is, it must pass through A. Therefore, if two circles, &c. Q. E. D.

Book III.

PROP. XII. THEOR.

If two circles touch each other externally, the straight line which joins their centres will 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 must pass through the point of contact A.

For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG: and because F is the centre of the circle ABC, AF

is equal to FC: Also because G is the centre of the circle ADE, AG is equal to GD. Therefore FA, AG are equal to FC,

B

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DG; wherefore the whole FG is greater than FA, AG; but it is also less, which is impossible: Therefore the a 20. 1. straight line which joins the points F, G, cannot pass otherwise than through the point of contact A; that is, it passes through A. Therefore, if two circles, &c. Q. E. D.

PROP. XIII. THEOR.

One circle cannot touch another in more points than one, whether it touch it on the inside or outside.

For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in the points B, D; join BD, and drawa GH, bisect- a 10. 11. 1.

Book III. ing BD at right angles: Therefore, because the points B, D

I

b 2. 3.

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are in the circumference of each of the circles, the straight line BD falls within each of them; and therefore their c Cor. 1. 3. centres are in the straight line GH which bisects BD at right angles: Therefore GH passes through the point of contact; 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 in the inside in more points than one.

d 11. 3.

K

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 ABC in the points A, C, and join AC: Therefore, because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them falls within the circle ACK: And the circle ACK is without the circle ABC; and therefore the straight line AC is also without ABC; but, because the points A, C are in the circumference of the circle ABC, the straight line AC is within the same circle, which is absurd: Therefore a circle can

b

B

not touch another on the outside in more than one point; and it has been shewn, that a circle cannot touch another on the inside in more than one point. Therefore, one circle, &c. Q. E. D.

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