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PROPOSITION XI.

THEORE M.
If two Circles touch each other on the Inside, and

the Centers be found, the Line joining their
Centers, will fall on the [Point of ] Contaxt

of those Circles.

ET two Circles ABC, ADE, touch one an

other inwardly in A, and let F be the Center of the Circle ABC, and G that of ADE. I say, 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.

Now because AG, GF, are greater than AF, * that is, than FH; take away F G, which is common, and 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 absurd. Wherefore a Line drawn thro' the Points F, G, will not fall out of the Point of Contact A, and so necessarily must fall in it; which was to be demonstrated,

* 20. I.

PROPOSITION XII.

THEORE M.
If two Circles touch one another on the Outside,

a Right Line joining their Centers will pass
thro' the [Point of ] Conta&t.
ET two Circles ABC, ADE, touch one ano-

ther outwardly in the Point A; and let F be the Center of the Circle ABC, and G that of ADE. I say, a Right Line drawn through the Center F, G, will pass thro' the Point of Contact A.

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

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

the

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

* 20. 1.

1

PROPOSITION XIII.

THE O R E M. One Circle cannot touch another in more Points than one, whether it be inwardly, or outwardly.

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OR, in the first place, if this be denied, let the

Circle ABDC, if possible, 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 tbis. 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 B G will be equal to GD. Therefore B G 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 possible, 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 assumed in the

Circumference of each of the Circles ABDC, ACK, | 2 of ibis a Right Line joining these two Points, will fall I

within either of the Circles. But it falls within the Circle ABDC, and without the Circle ACK, which is absurd. 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

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

PROPOSITION XIV.

THEOREM.
Equal Right Lines in a Circle are equally distant

from the Center; and Right Lines, which are
equally distant from the Center, are equal be-
tween themselves.

L

ET ABDC be a Circle, wherein are the equal

Right Lines AB, CD. I fay these Lines are equally distant 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 same. * 3 of this Wherefore AF is equal to FB, and so AB is double to AF. For the fame Reason 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 7 equal to the Square oft 47. I. AE. - For the Angle at F is a Right Angle; and the Squares of EG, and GC, are equal to the Square of EC, since 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. Also Lines in a Circle are I said to be equally distant from | Def. 4the 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 Construction being supposed, we demonstrate as above, that AB is double to AF, and

CD

† 47. 1.

CD to CG; and because AE is equal to EC, the
Square of A E will be equal to the Square of EC.
But the Squares of E F and FA, are † equal to the
Square of A E, 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 G C. But the Square of EG is equal to the
Square of EF; for EĞ'is equal to EF. Therefore
the Square of AF is equal to the Square of CG;
and so AF is equal to CG. But A B is double to
AF, and CD to CG. Therefore equal Right Lines
in a Circle are equally distant from the Center; and
Right Lines, which are equally distant from the Cena
ter, are equal between themselves; which was to be
demonstrated.

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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.

ET ABCD be a Circle, whose Diameter is

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the Diameter than F G. I say, 'AD is the greatest,
and B C is greater than F G.

For let the Perpendiculars EH, EK, be drawn
from the Center E to BC, FG. Now because BC
is nearer to the Center than F G, 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 E H is equal to EL, the Line BC * 14 of rbis. will be equal to MN*. And, since AE is equal

to EM, and D E to EN, AD will be equal to ME + 20. I. and EN. But ME and EN are f greater than MN:

And fo AD is greater than MN, and NM is equal
to BC. Therefore A D is greater than B C. And
fince the two Sides EM, EN, are equal to the two

Şides FE, EG, and the Angle MEN greater than $ 24. 10 the Angle FEG, the Base MN shall be I greater than the Base FG. But MN is equal to BC. There

fore

fore B C is greater than F G. And so the Diameter AD is the greatest, 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 romote; which was to be demonstrated.

PROPOSITION XVI.

THEOREM.
A Line drawn from the extreme [Point] of the Dia.

meter of a Circle at Right Angles to that Diame-
ter, Mall fall without the Circle; and between
the said Right Line, and the Circumference, no
other Line can be drawn; and the Angle of a Se-
micircle is greater than any Right-lin'd acute An-
gle; and the remaining Angle (withcut any Cir-
cumference] is less than any Right-lined Angle.

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ET ABC be a Circle, whose Center is D, and

Diameter AB. I say, 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 possible, within the Circle, as A C, and join D C.

Now because DA is equal to DC, the Angle DAC shall be * equal to the Angle ACD. But DAC is *

5. a Right Angle; therefore ACD is a Right Angle: And accordingly the Angles DAC, ACD, are equal to two Right Angles; which is absurd t. There- † 17. le fore a Right Line drawn from the Point A at Right Angles to BA, will not fall within the Circle ; and so likewise we prove, that it neither falls in the Circumference. Therefore it will necessarily fall without the same ; which now let be A E.

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 1-12. 1. drawn at Right Angles from the Center D to FA.

Now because AĞD is a Right Angle, and DAG is lefs than a Right Angle, D A will be greater than DG*, But DA is equal to DH. Therefore DH is * 19. 1.

greater

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