PROPOSITION XI.

THEOREM.

If two Circles touch each other on the Infide, and the Centers be found, the Line joining their Centers, will fall on the [Point of] Contact of thofe Circles.

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ET two Circles ABC, ADE, touch one another inwardly in A, and let F be the Center of the Circle ABC, and G that of ADE. I fay, a Right Line joining the Centers G and F, being produced, will fall in the Point A.

If this be denied, let the Right Line, joining FG, cut the Circle in D and H.

mon,

Now because AG, GF, are greater than AF, * that is, than FH; take away FG, which is comand 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 abfurd. Wherefore a Line drawn thro' the Points F, G, will not fall out of the Point of Contact A, and fo neceffarily muft fall in it; which was to be demonstrated,

if

PROPOSITION XII.

THEOREM.

two Circles touch one another on the Outfide, a Right Line joining their Centers will pass thro' the [Point of] Contact.

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ET two Circles ABC, ADE, touch one another outwardly in the Point A; and let F be the Center of the Circle ABC, and G that of ADE. I fay, a Right Line drawn through the Center F, G, will pass thro' the Point of Contact A.

For if it does not, let, if poffible, FCDG fall without it, and join FA, AG.

Now fince F is the Center of the Circle ABC, AF will be equal to FC. And becaufe G is the Center of

the

* 20. I.

*20. I.

the Circle ADE, AG will be equal to GD: but AF has been fhewn to be equal to FC; therefore FA, AG, are equal to FC, DG. And fo the whole FG is greater than FA, AG; and also lefs, * which is abfurd. Therefore a Right Line drawn from the Point F to G, will pass through the Point of Contact A; which was to be demonftrated.

PROPOSITION XIII. .

THEOREM.

One Circle cannot touch another in more Points than one, whether it be inwardly, or outwardly.

OR, in the first Place, if this be denied, let the Circle ABDC, if poffible, touch the Circle EBFD inwardly, in more Points than one, viz. in B, D.

And let G be the Center of the Circle ABDC, and H that of EBFD.

Then a Right Line drawn from the Point G to H, 11 of this. will + fall in the Points B and D. Let this Line be

BGHD. And because G is the Center of the Circle ABDC, the Line BG will be equal to GD. Therefore BG is greater than HD, and BH much greater than HD. Again, fince H is the Center of the Circle EBFD, the Line BH is equal to HD. But it has been proved to be much greater than it, which is abfurd. Therefore one Circle cannot touch another Circle inwardly in more Points than one.

Secondly, let the Circle ACK, if poffible, touch the Circle ABDC outwardly in more Points than one, viz. in A and C, and let A, C, be joined.

Now because two Points A, C, are affumed in the Circumference of each of the Circles ABDC, ACK, 2 of this a Right Line joining thefe two Points, will fall within either of the Circles. But it falls within the Circle ABDC, and without the Circle ACK, which is abfurd. Therefore one Circle cannot touch another Circle in more Points than one outwardly. But it has been proved, that one Circle cannot touch another Circle inwardly, [in more Points than one.] Wherefore one Circle cannot touch another in more

Points than one, whether it be inwardly or outwardly; which was to be demonftrated.

PROPOSITION XIV.

THEOREM.

Equal Right Lines in a Circle are equally distant from the Center; and Right Lines, which are equally diftant from the Center, are equal between themselves.

ET ABDC be a Circle, wherein are the equal Right Lines AB, CD. I fay these Lines are equally diftant from the Center of the Circle.

For let E be the Center of the Circle ABDC; from which let there be drawn EF, EG, perpendicular to AB, CD, and let AE, EC, be joined.

Then because a Right Line EF, drawn thro' the Center, cuts the Right Line AB, not drawn thro' the Center at Right Angles, it will bifect the fame. 3 of this Wherefore AF is equal to FB, and fo AB is double to AF. For the fame Reafon CD is double to CG, but AB is equal to CD. Therefore AF is equal to CG; and because AE is equal to EC, the Square of AE fhall be equal to the Square of EC. But the Squares of AF and FE are equal to the Square of† 47. 1. AE. For the Angle at F is a Right Angle; and the Squares of EG, and GC, are equal to the Square of EC, fince the Angle at G is a Right one. Therefore the Squares of AF and FE, are equal to the Squares of CG and GE. But the Square of AF is equal to the Square of CG; for AF is equal to CG. Therefore the Square of FE is equal to the Square of EG; and fo FE equal to EG. Alfo Lines in a Circle are faid to be equally diftant from Def. 4. the Center, when Perpendiculars drawn to them from of this. the Center are equal. Therefore AB, CD, are equally diftant from the Center.

But if AB, CD, are equally diftant from the Center, that is, if FE be equal to EG: I fay AB, is equal to CD.

For the fame Conftruction being fuppofed, we demonftrate as above, that AB is double to AF, and

CD

† 47. I.

*

† 20. I.

CD to CG; and because AE is equal to EC, the Square of AE will be equal to the Square of EC. But the Squares of EF and F A, are † equal to the Square of AE, and the Squares of EG, and GC, equal to the Square of EC. Therefore the Squares of EF and FA, are equal to the Squares of EG and GC. But the Square of EG is equal to the Square of EF; for E G is equal to EF. Therefore the Square of AF is equal to the Square of CG; and fo AF is equal to CG. But AB is double to AF, and CD to CG. Therefore equal Right Lines in a Circle are equally diftant from the Center; and Right Lines, which are equally distant from the Center, are equal between themselves; which was to be demonftrated.

PROPOSITION XV.

THEOREM.

A Diameter is the greatest Line in a Circle; and of all the other Lines therein, that which is nearest to the Center is greater than that more remote.

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ET ABCD be a Circle, whofe Diameter is AD, and center E; and let BC be nearer to the Diameter than FG. I fay, AD is the greatest, and BC is greater than F G.

For let the Perpendiculars EH, EK, be drawn from the Center E to B C, FG. Now because BC is nearer to the Center than FG, EK will be greater than EH. Let EL be equal to EH; draw LM through L at Right Angles to EK, which produce to N, and let EM, EN, EF, EG, be joined.

Then because EH is equal to EL, the Line BC 14 of this. will be equal to MN *. And, fince AE is equal to EM, and DE to EN, AD will be equal to ME and EN. But ME and EN are + greater than MN: And fo AD is greater than MN; and NM is equal to BC. Therefore AD is greater than BC. And fince the two Sides EM, EN, are equal to the two Sides FE, EG, and the Angle MEN greater than the Angle FEG, the Bafe MN fhall be greater than the Bafe F G. But MN is equal to BC. There

$24. T.

fore

fore BC is greater than F G. And fo the Diameter AD is the greateft, and B C is greater than F G. Wherefore the Diameter is the greatest Line in a Circle; and of all the other Lines therein, that which is nearest to the Center is greater than that more remote; which was to be demonftrated.

PROPOSITION XVI.

THEOREM.

A Line drawn from the extreme [Point] of the Diameter of a Circle at Right Angles to that Diame ter, fhall fall without the Circle; and between the faid Right Line, and the Circumference, no other Line can be drawn; and the Angle of a Semicircle is greater than any Right-lin❜d acute Angle; and the remaining Angle [without any Circumference] is less than any Right-lined Angle.

ET ABC be a Circle, whofe Center is D, and Diameter AB. I fay, a Right Line drawn from the Point A at Right Angles to AB, falls without the Circle.

For if it does not, let it fall, if poffible, within the Circle, as A C, and join DC.

*

Now because DA is equal to DC, the Angle DAC fhall be equal to the Angle ACD. But DAC is * 5. 1. a Right Angle; therefore ACD is a Right Angle: And accordingly the Angles DAC, ACD, are equal to two Right Angles; which is abfurd +. There- † 17. 1. fore a Right Line drawn from the Point A at Right Angles to BA, will not fall within the Circle; and fo likewise we prove, that it neither falls in the Circumference. Therefore it will neceffarily fall without the fame; which now let be AE.

Again, between the Right Line AE, and the Circumference CHA, no other Right Line can be drawn.

For if there can, let it be FA, and let ‡ DG be ‡-12. 1. drawn at Right Angles from the Center D to FA. Now because A ĞD, is a Right Angle, and DAG is lefs than a Right Angle, DA will be greater than DG*. But DA is equal to DH. Therefore DH is * 19 1.

2

greater