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Centre of the Circle A D E, A G will be equal to GD: But AF has been fhewn to be equal to FC; therefore FA and AG are equal to FC and DG. And fo the whole F G is greater than FA and AG; and alfo lefs, which is abfurd. Therefore, a Right* 20. 1. 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.

FO

OR, in the first place, if this be denied, let the Circle ABD C, if poffible, touch the Circle EBFD inwardly, in more Points than one, viz. in B and D.

And let G be the Centre of the Circle A B D C, and H that of EBF D.

Then a Right Line drawn from the Point G to H,

will fall in the Points B and D. Let this Line be† 11 of thie. + BGHD. And because G is the Centre of the Circle

ABDC, the Line B G will be equal to GD. Therefore B G is greater than HD, and B H much greater than HD. Again, fince H is the Centre of the Circle E B FD, the Line B H 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 A C K, if poffible, touch the Circle A B D C outwardly, in more Points than one, viz. in A and C ; and let A and C be joined.

Now, because two Points, A and C, are affume! in the Circumference of each of the Circles A B D C, AC K, a Right Line joining thefe two Points will fall within either of the Circles. But it falls within the ‡ 2 of this. Circle ABD C, and without the Circle A C K, 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

ones

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

PR

PROPOSITION XIV.
THEOREM.

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

LET ABDC be a Circle, wherein are the equal Right Lines A B, CD. I fay, thefe Lines are equally diftant from the Centre of the Circle.

For, let E be the Centre of the Circle A B DC; from which let there be drawn E F and EG, perpendicular to A B and CD; and let A E and E C be joined.

Then, because a Right Line E F, drawn thro' the Centre, cuts the Right Line A B, not drawn thro' the 3 of this. Centre, at Right Angles, it will bifect the fame.

† 47. I.

Wherefore A F is equal to F B, and fo A B is double to A F. For the fame Reafon CD is double to CG; but A B is equal to CD; therefore A F is equal to CG: And because A E is equal to E C, the Square of A E fhall be equal to the Square of E C. But the Squares of A F and F E are + equal to the Square of AE; for the Angle at F is a Right Angle: And the Squares of E G and G C are equal to the Square of EC, fince the Angle at G is a Right one. Therefore the Squares of AF and F E are equal to the Squares of CG and G E. But the Square of AF is equal to the Square of CG; for A F is equal to C G. Therefore the Square of F E is equal to the Square of E G; and fo F E equal to E G. Alfo Lines in a Circle are Def. 4 of faid to be equally diftant from the Centre, when Perpendiculars drawn to them from the Centre are equal. Therefore, A B and CD are equally diftant from the Centre.

this.

But if A B and C D are equally distant from the Centre, that is, if F E be equal to E G, I say, AB is equal to C D.

For, the fame Conftruction being fuppofed, we demonstrate, as above, that A B is double to A F, and

CD

CD to CG; and because A E is equal to E C, the
Square of A E will be equal to the Square of E C. But
the Squares of EF and F A are † equal to the Square † 47.1.
of AE; alfo the Squares of E G and G C are equal
+ to the Square of E C. Therefore the Squares of
EF and FA are equal to the Squares of EG and
GC. But the Square of E G is equal to the Square
of EF; for EG is equal to EF. Therefore the
Square of A F is equal to the Square of C G; and fo
AF is equal to C G. But A B is double to A F, and
CD to CG; whence A B is equal to CD. There-
fore, equal Right Lines in a Circle are equally diftant
from the Centre; and Right Lines, which are equally
diftant from the Centre, 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 Centre is greater than that more

remote.

LET ABCD be a Circle, whofe Diameter is AD,

and Centre E; and let BC be nearer to the Centre than F G. I fay, AD is the greateft, and BC is greater than F G.

For, let the Perpendiculars EH, EK, be drawn from the Centre E to B C, F G. Now, because B C is nearer to the Centre than F G, EK will be greater than EH. Let EL be equal to E H; draw L M through L at Right Angles to E K, which produce to N; and let E M, EN, EF, EG, be joined.

Then, because E H is equal to EL, the Line BC will be equal to MN. And fince A E is equal to 14 of this, E M, and D E to E N, AD will be equal to M E and EN. But ME and EN are + greater than MN : † 20. 1. And fo A D is greater than MN; and N M is equal to B C. Therefore A D is greater than B C. fince the two Sides E M, EN, are equal to the two Sides FE, EG, and the Angle MEN greater than the Angle FE G, the Bare M N fhall be greater ‡ 24. 1. than the Base F G. But M N is equal to B C. Therefore

And

† 17. 1.

+ 12, 1.

19.1.

Therefore BC is greater than F G. And fo the Diameter A D is the greateft, and B C is greater than FG. Wherefore, the Diameter is the greatest Line in a Circle; and of all the other Lines therein, that which is nearest to the Centre, 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 Diameter, hall fall without the Circle; and between the faid Right Line, and the Circumference, no other Right Line can be drawn ; and the Angle of a Semicircle is greater than any Right-lined acute Angle; and the remaining Angle [viz. without the Circumference] is lefs than any Right-lined Angle.

L

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

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

*

Now, becaufe DA is equal to D C, the Angle DAC fhall be equal to the Angle AC D. But DAC is a Right Angle; therefore A C D is a Right Angle: And accordingly the Angles DAC, ACD, are equal to two Right Angles; which is abfurd +. Therefore a Right Line drawn from the Point A at Right Angles to B A, will not fall within the Circle; and fo likewife we prove, that it neither falls in the Circumference. Therefore, it will neceffarily fall without the fame; which now let be A E.

Again, between the Right Line A E and the Circumference CH A, no other Right Line can be drawn. For, if there can, let it be F A, and let ‡ DG be drawn from the Centre D, at Right Angles to F A.

Now, because A G D is a Right Angle, and DAG is less than a Right Angle, D A will be greater than DG. But DA is equal to D H. Therefore D His greater

greater than DG, the lefs than the greater; which is abfurd. Wherefore, no Right Line can be drawn between A E, and the Circumference AHC. I fay, moreover, that the Angle of the Semicircle, contained under the Right Line B A, and the Circumference CHA, is greater than any Right-lined acute Angle; and the remaining Angle contained under the Circumference C H A, and the Right Line A E, 'is lefs than any Right-lined Angle.

For if any Right lined acute Angle be greater than the Angle contained under the Right Line B A, and the Circumference C HA; or if any Right-lin'd An-. gle be less than that contained under the Circumference CHA, and the Right Line A E; then a Right Line may be drawn between the Circumference C H A, and the Right Line A E, making an Angle (contained under Right Lines) greater than that contained under the Right Line B A, and the Circumference C HA, and less than that contained under the Circumference CHA, and the Right Line A E. But fuch a Right Line cannot be drawn, from what has been proved. Therefore, no Right-lin'd acute Angle is greater than the Angle contained under the Right Line B A, and the Circumference CHA; nor less than the Angle contained under the Circumference CHA, and the Right Line A E; which was to be demonftrated.

Coroll. From hence it is manifeft, that a Right Line, drawn at Right Angles, on the End of the Diameter of a Circle, touches the Circle, and that in one Point only; because, if it should meet it in two Points, it would fall within the fame; * as has been 2 of this. demonftrated.

PROPOSITION XVII.

PROBLEM.

To draw a Right Line from a given Point, that
fhall touch a given Circle.

L ETA be the Point given, and B CD the Circle.
It is required to draw a Right Line from the Point
A, that fhall touch the given Circle B C D.

G

Let

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