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A

less than D H.

M

of the circle is DA, which paffes thro' the centre, and amongst the reft that will always be the greater which is the nearer to that paffing thro' the centre, viz. DE greater than D F, and D F than D C. But of those lines which fall upon the convex part HLKG of the circumference D G, which lies between the diameter and the point D, is the leaft, and that line which is nearer to this, is always lefs than that which is more remote, viz. D K less than DL, and DL

For [by 1.
3.] find the centre м of the circle A B C,
and join ME, MF, MC, MK, ML, M H.

Now because A M is equal to E M, let M D, which is common, be added; and then A D is equal to EM, MD. But [by 20. 1.] E M, M D are greater than ED: therefore AD is greater than E D. Again, because M E is equal to MF, and MD is common, EM, MD will be equal to MF, MD, and the angle EMD is greater than the angle F MD; therefore the bafe E D [by 24. 1.] will be greater than the base FD. After the fame manner we prove, that FD is greater than CD; therefore DA is the greatest of all thofe lines falling from D upon the circle; DE is greater than DF, and DF greater than D C.

Again, because M K, KD [by 20. 1.] are greater than MD, and MG is equal to M K, the remainder K D will be greater than the remainder GD: wherefore G D is less than KD, and fo it is the leaft of them all; and because two right lines MK, KD ftand upon one fide MD of the triangle MLD within it, MK, KD [by 21. 1.] will be lefs than ML, LD; but м K is equal to ML. Therefore the remainder D K is lefs than the remainder DL. In like manner we demonftrate that D L is less than DH, therefore D G is the least of these right lines; D K is lefs than DL, and D L lefs than D H.

I fay alfo, that only two equal right lines fall from D upon the circumference of the circle, the one on one fide, and the other on the other fide of the leaft line; for [by 23. 1.] at the given point м in the right line D make the

angle

angle D M B with it equal to the angle K M D, and join DB ; then because M K is equal to M B, and M D is common, the two right lines K M, M D, are equal to the two right lines BM, MD, each to each, and the angle K M D is equal to the angle B MD: therefore [by 4. 1.] the bafe K D is equal to the bafe D B. I fay alfo, that no other right line can fall from D, upon the circumference of the circle equal to D K. For if there can, let it be D N. And because D K is equal to D N, and D B equal to D K; the right line D B will also be equal to DN, that is, the line which is the nearer equal to that which is more remote. Which has been demonftrated to be impoffible,

Or thus: Join NM, and because K M is equal to MN, and M D is common, and the base D K is equal to the bafe DN; the angle K мD [by 8. 1.] will be equal to the angle DM N, but the angle K M D is equal to the angle в MD. Therefore the angle B M D will be equal to the angle N M D, the leffer equal to the greater: which is impoffible. Wherefore there cannot fall from the point D upon the circumference of the circle A B C, more than two equal right lines, the one on one fide the least line G D, and the other on the other fide of it.

If therefore any point be taken without a circle, &c. Which was to be demonftrated.

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If fome point be taken within a circle, and from it more than two equal right lines be drawn to the circumference, that affumed point is the centre of the circle.

Let the circle be A B C, and the point D taken within it, from which more than two right lines drawn to the circumference are equal to one another, viz. the three right lines DA, DB, DC. I fay, the point D is the centre of the circle A B C.

For, join A B, BC, and [by 10. 1.] divide them into equal parts at the points E, F; join E D, D F, which produce to the points G, K, H, L.

Then because A E is equal to E B, and ED is common; the two fides A E, ED will be equal to the two fides BF,

I 4

ED,

Б

F

E

D

K

H

G

ED, and the base DA is equal to the base D B. Therefore the angle A E D will be equal to the angle BED [by 8. 1.] And fo [by 10. def. 1.] the angles AED, BED are both right angles; wherefore fince GK bifects A B, it will cut the fame at right angles; and because if one right line in a circle bifects a right line in it

at right angles, the centre of the circle will fall [by cor. I. 3.] in the bifecting line; the centre of the circle A B C will fall in the right line GK. For the fame reason, the centre of the circle A B C will fall in the right line HL. But the right lines GK, HL have no point but D common to them both. Therefore D is the centre of the circle А В С.

If therefore fome point be taken within a circle, and from it more than two equal right lines be drawn to the circumference of the circle, that affumed point is the centre of the circle. Which was to be demonstrated.

F

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For take fome point D, within the circle A B C, and let more than two right lines DA, D B, DC, fall upon the circumference of the circle. I fav, the affumed point D is the centre of the circle A BC.

DE

A

B

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For if not, let, if poffible, the point E be the centre, and having joined D E produce it to F G. Then F G the diameter of the circle And fo because fome point D is taken in the diameter F G of the circle A B C, not being its centre [by 7. 3.] the greatest line falling upon the circumference will be DG; but DC is greater than D B, and D B than D A. But they are equal too; which cannot be. Wherefore E is not the centre of the circle A B C. After the fame manner we demonftrate that no other point befides D can be the centre of the circle. Therefore D is the centre of the circle A B C.

PROP.

PROP. X. THEOR.

One circle cannot cut another in more points than

two.

For if poffible, let one circle A B C cut another circle DE F in more points than two, viz. in the points B, G, H, and join the right lines B G,

BH, which bisect in K, L,

and draw KC, LM, from

K, L at right angles to B G, BH, and produce them to the points A and E.

Then because in the circle A B C, the right line A C bifects the right line в H, at right angles; the centre of the circle ABC [by cor.

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1. 3.] will fall in AC. Again, because in the fame circle A B C the right line N x cuts the right line B G into two equal parts, and at right angles too, the centre of the circle ABC will fall in the right line N x. But it has been proved that the centre is in A C too; and the right lines A C, NX will not meet one another in any other point but o; therefore o is the centre of the circle A B C. In like manner we prove, that o is the centre of the circle D E F : therefore o is the centre of both the circles A B C, DEF cuting each other, which [by 5. 3.] is impoffible.

Therefore one circle cannot cut another in more points than two. Which was to be demonstrated.

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point K [by 9. 3.] will be the centre of the circle DEF, but K is the centre of the circle. A B C. Therefore K will be the centre of two circles that interfect one another, Which [by 5. 3.] is impoffible.

Wherefore one circle cannot cut another in more points than two. Which was to be demonftrated.

PROP. XI. THEOR,

If two circles touch one another inwardly, and their centres be taken, the right line joining their centres being produced, will fall in the point wherein those circles touch one another.

For let two circles A B C, A D E, touch one another inwardly in the point A, and let F be the centre of the circle A B C, and G the centre of the circle ADE: I say, the right line joining the points G and being produced will fall in the point A.

For if not, let, if poffible, the right line F G D H, joining the centres, fall without the point of contact, and join A F, AG.

H

Then because AG, GF are [by 20. 1.] greater than A F, that is than F H (for F A is equal to F H, both being drawn from the fame centre). If F G which is common be taken away, the remainder AG is greater than the remainder GH. But A G is equal to GD; therefore GD is greater than GH, the Cleffer than the greater; which is im

G

E

B

poffible. Therefore a right line joining the points FG, being continued, will not fall without the point of contact A; and fo neceffarily muft fall in it.

If therefore two circles touch one another inwardly, and their centres be taken, the right line joining their centres will fall in the point of contact of thofe circles. Which was to be demonstrated.

Otherwife:

*But let it fall without the point of contact, as GFC, continue out C F G to the point H, and join AG, AF.

*This demonfiration fcarcely differs from the former, fo that it feems not to be Euclid's.

Then

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