If not, let it pass otherwise, if possible, as FCDG, and join FA, AG.
And because F is the center of the circle ABC,
FA is equal to FC:

also, because G is the center of the circle ADE,
GA is equal to GD:

therefore FA, AG are equal to FC, DG; (ax. 2.) wherefore the whole FG is greater than FA, AG: but FG is less than FA, AG; (I. 20.) which is impossible: therefore the straight line which joins the points F, G, cannot pass otherwise than through A the point of contact,

that is, FG must pass through the point A.

Therefore, if two circles, &c.

Q. E. D.

PROPOSITION XIII. THEOREM.

One circle cannot touch another in more points than in one, whether it touches it on the inside or outside.

For, if it be possible, let the circle EBF touch the circle ABC in more points than in one.

and first on the inside, in the points B, D.

Join BD, and draw GH bisecting BD at right angles. (I. 11.) Because the points B, D are in the circumferences of each of the circles, therefore the straight line BD falls within each of them; (III. 2.) therefore their centers are in the straight line GH which bisects BD at right angles; (III. 1. Cor.)

therefore GH passes through the point of contact: (III. 11.)
but it does not pass through it,

because the points B, D are without the straight line GH;
which is absurd:

therefore one circle cannot touch another on the inside in more points

than in one.

Nor can two circles touch one another on the outside in more than

in one point.

For, if it be possible,

let the circle ACK touch the circle ABC in the points A, C;

join AC.

B

A

Because the two points A, C are in the circumference of the circle ACK,

therefore the straight line AC which joins them, falls within the circle ACK: (III. 2.)

but the circle ACK is without the circle ABC; (hyp.) therefore the straight line AC is without this last circle: but, because the points A, C are in the circumference of the circle ABC. the straight line AC must be within the same circle, (III. 2.)

which is absurd;

therefore one circle cannot touch another on the outside in more than in one point:

and it has been shewn, that they cannot touch on the inside in more points than in one.

Therefore, one circle, &c.

Q. E.D.

PROPOSITION XIV. THEOREM.

Equal straight lines in a circle are equally distant from the center; and conversely, those which are equally distant from the center, are equal to one another.

Let the straight lines AB, CD, in the circle ABDC, be equal to one another.

Then AB and CD shall be equally distant from the center.

Take E the center of the circle ABDC, (III. 1.)

from E draw EF, EG perpendiculars to AB, CD, (1. 12.) and join EA, EC.

Then, because the straight line EF passing through the center, cuts AB, which does not pass through the center, at right angles; EF bisects AB in the point F: (III. 3.)

therefore AF is equal to FB, and AB double of AF.
For the same reason CD is double of CG:
but AB is equal to CD: (hyp.)
therefore AF is equal to CG. (ax. 7.)
And because AE is equal to EC, (ì. def. 15.)
the square on AE is equal to the square on EC:

but the squares on AF, FE are equal to the square on AE, (1. 47.) because the angle AFE is a right angle ;

and for the same reason, the squares on EG, GC are equal to the square on EC; therefore the squares on AF, FE are equal to the squares on CG, GE: (ax. 1.)

but the square on AF is equal to the square on CG,

because AF is equal to CG;

therefore the remaining square on EF is equal to the remaining square on EG, (ax. 3.)

and the straight line EF is therefore equal to EG:

but straight lines in a circle are said to be equally distant from the center, when the perpendiculars drawn to them from the center are equal: (III. def. 4.)

therefore AB, CD are equally distant from the center. Conversely, let the straight lines AB, CD be equally distant from the center, (III. def. 4.)

that is, let FE be equal to EG;

then AB shall be equal to CD.

For the same construction being made,
it may, as before, be demonstrated,

that AB is double of AF, and CD double of CG,

and that the squares on FE, AF are equal to the squares on EG, GC: but the square on FE is equal to the square on EG, because FE is equal to EG; (hyp.)

therefore the remaining square on AF is equal to the remaining square on CG: (ax. 3.)

CD double of CG;

and the straight line AF is therefore equal to CG: but AB was shewn to be double of AF, and wherefore AB is equal to CD. Therefore equal straight lines, &c.

PROPOSITION XV.

(ax. 6.)

Q. E. D.

THEOREM.

The diameter is the greatest straight line in a circle; and of the rest, that which is nearer to the center is always greater than one more remote: and conversely the greater is nearer to the center than the less.

Let ABCD be a circle, of which the diameter is AD, and the center E; and let BC be nearer to the center than FG.

Then AD shall be greater than any straight line BC, which is not a diameter, and BC shall be greater than FG.

A B

F

KE

D

From E draw EH, perpendicular to BC, and EK to FG, (1. 12.) and join EB, EC, EF.

And because AE is equal to EB, and ED to EC, (1. def. 15.) therefore AD is equal to EB, EC: (ax. 2.) but EB, EC are greater than BC; (1. 20.) wherefore also AD is greater than BC.

And, because BC is nearer to the center than FG, (hyp.)
therefore EH is less than EK: (III. def. 5.)

but, as was demonstrated in the preceding proposition,
BC is double of BH, and FG double of FK,

and the squares on EH, HB are equal to the squares on EK, KF: but the square on EH is less than the square on EK, because EH is less than EK;

therefore the square on BH is greater than the square on FK, and the straight line BH greater than FK,

and therefore BC is greater than FG.

Next, let BC be greater than FG;

then BC shall be nearer to the center than FG, that is, the same construction being made, EH shall be less than EK. (III. def. 5.) Because BC is greater than FG,

BH likewise is greater than KF:

and the squares on BH, HE are equal to the squares on FK, KE, of which the square on BH is greater than the square on FK, because BH is greater than FK:

therefore the square on EH is less than the square on EK,
and the straight line EH less than ĒK:

and therefore BC is nearer to the center than FG. (III. def. 5.) Wherefore the diameter, &c.

Q.E. D.

PROPOSITION XVI. THEOREM.

The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn from the extremity between that straight line and the circumference, so as not to cut the circle: or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at right angles to it, as not to cut the circle. Let ABC be a circle, the center of which is D, and the diameter AB. Then the straight line drawn at right angles to AB from its extremity A, shall fall without the circle.

For, if it does not, let it fall, if possible, within the circle, as AC; and draw DC to the point C, where it meets the circumference. And because DA is equal to DC, (1. def. 15.) the angle DAC is equal to the angle ACD: (1. 5.) but DAC is a right angle; (hyp.)

therefore ACD is a right angle;

and therefore the angles DAC, ACD are equal to two right angles; which is impossible: (I. 17.)

therefore the straight line drawn from A at right angles to BA, does not fall within the circle.

In the same manner it may be demonstrated, that it does not fall upon the circumference; therefore it must fall without the circle, as AE.

Also, between the straight line AE and the circumference, no straight line can be drawn from the point A which does not cut the circle. For, if possible, let AF fall between them,

FE

H

and from the point D, let DG be drawn perpendicular to AF, (1. 12.) and let it meet the circumference in H.

And because AGD is a right angle, and DAG less than a right angle, (1. 17.) therefore DA is greater than DG: (1. 19.) but DA is equal to DH; (1. def. 15.) therefore DH is greater than DG,

the less than the greater, which is impossible:

therefore no straight line can be drawn from the point A, between AE and the circumference, which does not cut the circle :

or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point A, or however small an angle it makes with AE, the circumference must pass between that straight line and the perpendicular AE. Q.E.D.

COR. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it touches the circle; (III. def. 2.) and that it touches it only in one point, because, if it did meet the circle in two, it would fall within it. (111. 2.) "Also, it is evident, that there can be but one straight line which touches the circle in the same point."

PROPOSITION XVII. PROBLEM.

To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle.

First, let A be a given point without the given circle BCD; it is required to draw a straight line from A which shall touch the circle.

CE

Find the center E of the circle, (III. 1.) and join AE;

and from the center E, at the distance EA, describe the circle AFG; from the point D draw DF at right angles to EA, (1. 11.) meeting the circumference of the circle AFG in F;

and join EBF, AB.