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21. 1.

+15 Def. 1. † 5 Ax.

† 23. 1.

+ Constr.

4.1.

than the remainder GD, that is, GD is less than KD: and because MLD is a triangle, and from the points M, D, the extremities of its side MD, the straight lines MK, DK are drawn to the point K within the triangle, therefore MK, KD, are less than ML, LD: but 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 less 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 line. At the point M, in the straight line MD, make+ the angle DMB 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, each to each; and the angle KMD is equal to the angle BMD; therefore the base DK is equal 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, a line nearer to the least is equal to one more remote, which has been proved to be impossible. If therefore any point, &c.

Q. E. D.

PROP. IX. THEOR.

*

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 shall be 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

†17 Def. 1. diameter† of the circle ABC: and be

7. S.

↑ Hyp.

cause in FG, the diameter of the circle
ABC, there is taken the point D, which
is not the centre, DG is the greatest line
from it to the circumference, and DC is
greater than DB, and DB greater than

*

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DA: but they are likewise + equal, which is impossible:

In

therefore E is not the centre of the circle ABC.
like manner it may be demonstrated, that no other point
but D is the centre; D therefore is the centre. Where-
fore, if a point be taken, &c. Q. E. D.

PROP. X. THEOR.

One circumference of a circle cannot cut another in more than two points.

E

B D H

K

F

† 3.3.

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: then because K is the centre of the circle ABC, therefore KB, KG, KF are all equal to each other: and be- +15 Def. 1. cause 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, therefore the point K is the centre of the circle DEF: but K is 9.3. also the centre of the circle ABC; therefore the same + Constr. point is the centre of two circles that cut one another, which is impossible*. Therefore one circumference of 5.3. a 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, shall pass through the point A.

H

DGF

B

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For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG. Then, because two sides of a triangle are together greater than the third side, therefore FG, GA are greater than FA: but FA is equal to FH; there fore FG, GA are greater than FH: take away the common part FG; therefore the remainder AG is greater† † 5 Ax.

† 20. 1. +15 Def. 1.

+15 Def. 1. 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, being produced, cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c.

† 2 Ax.

* 20.1.

See N.

PROP. XII. THEOR.

Q. E. D.

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.

B

E

For, if not, let it pass otherwise, if possible, as
FCDG, and join FA, AG. And because F is the cen-
tre of the circle ABC, FA is equal
to FC: also, because G is the centre
of the circle ADE, GA is equal to
GD: therefore FA, AG are equal+
to FC, DG; wherefore the whole
FG is greater than FA, AG: but it

*

is also less; which is impossible: therefore the straight
line which joins the points F, G, cannot 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.

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 EBF touch the
circle ABC in more points than one, and first on the

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*

2. 3.

* Cor. 1. 3.

inside, in the points B, D: join BD, and draw* GH 10. 11. 1. bisecting BD at right angles: therefore, because the points B, D, are in the circumference of each of the circles, the straight line BD falls within each of them; therefore their 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 on the inside in more points than

one.

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: 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: but the circle ACK is without+ the circle ABC; 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 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 they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q. E. D.

PROP. XIV. THEOR.

B

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 shall be equally distant from the centre.

11. 3.

2.3.

+ Hyp.

2.3.

† 12. 1.

Take+ E the centre of the circle ABDC, and from itt †1.3. draw EF, EG, perpendiculars to AB, CD, and join EA, EC. 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* it: therefore AF is

B

E

* 3.3.

F

+ Hyp.

+7 Ax.

* 47.1.

† 1 Ax.

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*

equal to FB, and AB double of AF: for the same rea son CD is double of CG: but AB is equal + to CD; therefore AF is equal to CG. And because AE is +15 Def. 1. equal + to EC, the square of AE is equal to the square of EC: but the squares of AF, FE, are equal to 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 equal to the squares of CG, GE: but the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of EF is equal to the remaining square of EG, and the straight line EF is therefore equal to EG: but straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre * 4 Def. 3. are* equal: therefore AB, CD are equally distant from the centre.

† 3 Ax.

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Next, let the straight lines AB, CD be equally disDef. 3. tant from the centre, that is + let FE be equal to EG; AB shall be 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: but 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: but AB was shewn to be double of AF, and CD double of CG; wherefore AB is equal + to CD. Therefore equal straight lines, &c. Q. E. D.

+ Hyp. † 3 Ax.

+ 6 Ax.

See N.

† 12. 1.

PROP. XV. THEOR.

The diameter is the greatest straight line in a circle; 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 the diameter is AD, and the centre E; and let BC be nearer to the centre than FG: AD shall be greater than any straight line BC, which is not a diameter, and BC shall be greater than FG.

B

H

From the centre draw+ EH, EK perpendiculars to

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