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the Centers be found, the Line joining their
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.
a Right Line joining their Centers will pass
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 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.
THE O R E M. 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 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 than one, whether it be inwardly or outwardly; which was to be demonstrated.
from the Center; and Right Lines, which are
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
† 47. 1.
CD to CG; and because AE is equal to EC, the
all the other Lines therein, that which is nearest
ET ABCD be a Circle, whose Diameter is
the Diameter than F G. I say, 'AD is the greatest,
For let the Perpendiculars EH, EK, be drawn
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
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 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.
meter of a Circle at Right Angles to that Diame-
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.