Then because A G, GF are greater than AF, and AF is equal to CF, that is, to F H; if F G which is common be taken away, the remainder AG will be greater than the remainder GH; that is, GD is greater than GH, the less than the greater; which cannot be. In like manner we prove the fame abfurdity will follow, if the centre of the greater circle be without the lefs.

PROP. XII. THEOR.

If two circles touch one another outwardly, the right line joining their centres will pass thro' the point wherein the circles touch one another.

B

E

For let two circles AB C, AD E, touch one another outwardly in the point A. Let F be the centre of the circle A B C, and G that of the circle ADE: I fay, the right line drawn from the point F to G, will pass thro' the point of contact A.

For if not, let it fall with

A

out that point, as F C DG, and join a F, A G.

G

Then because F is the centre of the circle ABC, F C will be equal to FA. Again, because G is the centre of the circle A DE, AG will be equal to G D, but it has been proved that F A is equal to F c too. Therefore F A, AG are equal to FC, DG. Confequently the whole FG is greater than FA, AG. But [by 20. 1.] it is leffer too; which cannot be. Therefore the right line drawn from F to G cannot pafs otherwife than thro' the point A of contact: wherefore it neceffarily paffes thro' that point.

If therefore two circles touch one another outwardly, the right line joining their centres will pafs thro' the point of contact of thofe two circles. Which was to be demonftrated.

PROP. XIII. THEOR.

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

For firft, if poffible, let the circle A B D C touch the circle EBFD inwardly in more points than one, viz. in

BD,

BD, and [by 1. 3.] find G the centre of the circle ́À B DC, and H the centre of the circle E B F D.

Then a right line drawn thro' the points G, H will pafs thro' the points B D. Let this be BGH D, and because G is the centre of the circle AB D C, B G will be equal to G D.

K

E

B

GH

F

Therefore B G is greater than HD, and BH much greater than HD; Again, because H is the centre of the circle E B F D, BH is equal to HD; but it has been proved to be much greater than it; which is imD poffible. Therefore one circle will not touch another inwardly in more points than one.

Nor fecondly, I fay, one circle will not touch another outwardly in more points than one. For if poffible, let the circle ACK touch the circle ABDC outwardly in more points than one, viz. in A, C, and draw

A C.

Then because two points A, C, are taken in the circumference of the circles A B D C, A CK; the right line joining them [by 2. 3.] falls within them. But [by 3. def. 3.] the fame line which falls within the circle A BDC will fall without the circle A C K, which is abfurd. Therefore one circle cannot touch another outwardly in more points than one; but it has been proved, that one circle cannot touch another inwardly in more points than one.

Therefore one circle cannot touch another in more points than one, whether it be inwardly or outwardly. Which was to be demonstrated.

PROP. XIV.

THEOR...

In a circle equal right lines are equally distant from the centre; and thofe right lines which are equally diftant from the centre, are equal to one another.

Let there be a circle A B D C, and let the right lines AB, CD in it be equal: I fay, they are equally diftant from the centre.

For find E the centre of the circle A B D C, and from it draw IF, EG perpendicular to AB, CD, and join AE, EC,

Then

B

D

Then because the right line EF drawn from the centre cuts the right line A B not drawn from the centre at right angles [by 3.3.]it will divide it into two equal parts; wherefore AF is equal to

FB; and fo A B will be double to

E

F

G

AF. Also, for the fame reason

Α.

CD is double to CG; but AB is equal to CD; whereof AF is And because alfo equal to C G. A E is equal to E C, the fquare of A E will be equal to the fquare of E C. But [by 47. 1.] the fquares of AF, FE are equal to the fquare of AE: For the angle F is a right angle; alfo the fquares of E G, G C are equal to the square of EC, fince the angle G is a right one: therefore the fquares of A F, FE are equal to the fquares of c G, GE: but the fquare of A F is equal to the fquare of C G, for A F is equal to c G. Therefore the remaining square of F will be equal to the remaining square of E G; and fo F E is equal to E G. But in a circle right lines are faid [by 4. def. 3.] to be equally diftant from the centre, when the perpendiculars drawn to them from the centre are equal: therefore A B, C D are equally distant from the centre.

Now let A B C D be equally diftant from the centre, that is, let F E be equal to EG: I fay, AB will be equal

to C D.

For the conftruction remaining the fame, we demonftrate as above, that A B is double to A F, and C D double to C G and because A E is equal to E C, the fquare of A B will be equal to the fquare of E C. But [by 47. 1.] the fquares of E F, F A are equal to the square of A E, and the fquares of EG, G C to the fquare of E c. Therefore the fquares of EF, FA are equal to the fquares of EG, GC; but the fquare of EG is equal to the fquare of E F, fince EF is equal to EG. Therefore the remaining fquare of A F will be equal to the remaining square of CG; wherefore AF is equal to CG, but AB is double to AF, and CD double to Ć G. Therefore A B will be equal to C D.

Wherefore in a circle equal right lines are equally diftant from the centre, and those right lines which are equally distant from the centre are equal to one another. Which was to be demonstrated.

PROP.

PROP. XV. THEOR.

The greateft right line in a circle is a diameter, and of all other right lines in it, that which is nearer to the centre is greater than that which is more

remote.

Let A B C D be a circle, whofe diameter is A D, and centre E, and let B c be nearer to the centre E than F G. I fay, A D is the greatest right line, and B C is greater than

FG.

MAB

For from the centre E draw the right lines E H, E K perpendicular to BC, F G. Then because B C is nearer to the centre than FG [by 5. def. 3.] EK will be greater than E H, make E L equal to E H, and through L draw LM at right angles to E K; which produce to N, and join EM, EN, EF, EG.

G

NDC

H

Then because EH is equal to EL; the line BC [by 14. 3.] will be equal to M N. Again, because AE is equal to E M, and ED to EN: A D will be equal to M E, E N; but M E, EN are greater than MN; therefore A D is greater than M N. But M N is equal to B C: therefore A D is greater than в C. And because the two fides ME, E N are equal to the two fides FE, EG, and the angle MEN is greater than the angle F EG; the base M N [by 24. 1.] will be greater than the base FG. But MN has been proved to be equal to BC: Wherefore the base B C is greater than the base F G. Therefore the diameter A D is the greatest of all these right lines, and B C is greater than F G.

Wherefore the greateft right line drawn in a circle is a diameter; and of all other right lines in it, that which is nearer to the centre is greater than that which is more reWhich was to be demonftrated.

mote.

PROP. XVI. THEOR.

A right line drawn from the end of the diameter of a circle, at right angles to that diameter, will fall without the circle; and no other right line can be

drawn

drawn between that Jame line and the circumference of the circle; the angle of a femicircle is greater than any right-lined acute angle; and the remaining angle (made by that line and the circumference) is less than any right-lined angle 8.

Let A B C be a circle, whofe centre is D, and diameter AB: I fay, the right line drawn from the point A, at right angles to A B, will fall without the circle.

For if not, let it fall within if poffible, as a c, and join

DC.

Then because DA is equal to D C, the angle DAC [by 5. 1.] will be equal to the angle A DC; but DAC is a right angle; therefore A C D is a right angle too; and fo the angles DAC, A C D are equal to two right angles, which [by 17. 1.] cannot be. Therefore a right line drawn from the point A, at right angles to BA, does not fall within the circle; after the fame manner we demonstrate that it cannot fall in the circumference; therefore it must neceffarily fall without the circle, as AE does.

I say 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 draw D G from the point D perpendicular to F A.

Then because the angle AGD is a right angle, and DAG

is less than a right angle, A D
[by 19. 1.] will be greater
than D G. But DA is equal
to DH: therefore D His
greater than DG, the less
than the greater; which is
impoffible. Therefore no right
line can be drawn between B
AE and the circumference
of the circle.

I fay moreover, the angle of the femicircle, which

E

F

G

ΙΑ

D

is 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 HA, and the right line A E, is less than any right-lined acute angle.

For