cumference DEF in more than two points, viz. in B, G, F; take the centre K of the circle ABC, and join KB, KG, KF: And because within the circle Def there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, the point K is (III. 9.) the centre of the circle DEF: But K is also the centre of the circle ABC: therefore the same point is

E

K

C

H

the centre of two circles that cut one another, which is impossible (III. 5.). Therefore one circumference of a circle cannot cut another in more than two points. Q.E.D.

PROP. XI. THEOR.

If two circles touch each other internally, the straight line which joins their centres being produced shall pass through the point of contact.

A

Let the two circles ABC, ADE, touch each other internally in the point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE: The straight line which joins the centres F, G, being produced, passes through the point A.

For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG, GF are greater (1. 20.) than FA, that is, than FH, for FA is equal to FH, both being from the same centre; take away the common part FG; therefore the remainder AG is greater than

H

D

E

the remainder GH: But AG is equal to GD; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c. Q. E. D.

PROP. XII. THEOR.

If two circles touch each other externally, the straight line which joins their centres shall pass through the point of

contact.

Let the two circles ABC, ADE touch each other externally in the point a: and let F be the centre of the circle ABC, and G the centre of ADE: The straight line which joins the points F, G shall pass through the point of contact A.

B

E

For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG: And because F is the centre of the circle ABC, AF is equal to FC: Also because G is the centre of the circle ADE, AG is equal to GD: Therefore FA, AG are equal to FC, DG; wherefore the whole FG is greater than FA, AG; But it is also less (1. 20.); which is impossible: Therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A; that is, it must pass through it. Therefore, if two circles, &c. Q. E. D.

F

PROP. XIII. THEOR.

D

One circle cannot touch another in more points than 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 one, and first on the inside, in the points B, D; join BD, and draw (1. 10, 11.) GH bisecting BD at right angles: Therefore, because the points B, D are in

the circumference of each of the circles, the straight line BD falls within each (ш. 2.) of them: And their centres are

(II. 1. Cor.) in the straight line GH which bisects BD at right angles: 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 one.

:

K

Nor can two circles touch one another on the outside in more than one point: For, if it be possible, let the circle ACK touch the circle ABC in the points A, C, and join AC: Therefore, because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them shall fall within (II. 2.) the circle ACK: And the circle ACK is without the circle ABQ and 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 (II. 2.) the same circle, which is absurd: There- B fore one circle cannot touch another on the

outside in more than one point: And it has been shown, that they cannot touch on the inside in more points than one : Therefore, one circle, &c. Q. E. D.

PROP. XIV. THEOR.

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

Let the straight lines AB, CD, in the circle ABDC, be equal to one another; they are equally distant from the centre.

C

Take E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD: Then, because the straight line EF passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects (III. 3.) it: Wherefore AF is equal to FB, and AB double of AF. For the same reason, CD is double of CG: And AB is equal to CD (Hyp.); therefore, AF is equal to CG: F And because AE is equal to EC, the square of AE is equal to the square of EC: But the squares of AF, FE are equal (1. 47.) to

B

E

G

the square of AE, because the angle AFE is a right angle: and, for the like reason, the squares of EG, GC are equal to the square of EC: Therefore the squares of AF, FE are equal to the squares of CG, GE,. of which the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line EF is therefore equal to EG: but straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal (III. Def. 4.): Therefore AB, CD are equally distant from the centre.

Next, if the straight lines A B, CD be equally distant from the centre, that is, if FE be equal to EG; AB is 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 of EF, FA are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG: But AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D.

PROP. XV. THEOR.

The diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the less.

Let ABCD be a circle, of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG; AD is greater than any straight line BC which is not a diameter, and BC greater than FG.

From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; and because AE is equal to EB, and ED to EC: AD is equal to EB, EC: but EB, EC, are greater (1. 20.) than BC; wherefore also AD is greater than BC.

F

Κ

A B

D

H

C

And because BC is nearer to the centre than FG, EH is less (II. Def. 5.) than EK: But, as was demonstrated in the

preceding, BC is double of BH, and FG double of FK, and the squares of EH, вH are equal to the squares of E-K, KF, of which the square of EH is less than the square of EK, because EH is less than EK; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK; and therefore BC is greater than FG.

Next, let BC be greater than FG; BC is nearer to the centre than FG, that is, the same construction being made, EH is less than EK: because BC is greater than FG, BH likewise is greater than KF: And the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK; therefore the square of EH is less than the square of EK, and the straight line EH less than EK. Wherefore the diameter, &c. QE. D.

PROP. XVI. THEOR.

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 between that straight line and the circumference from the extremity, 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 centre of which is D, and the diameter AB: 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 pos

sible, within the circle, as AC, and draw
DC to the point c where it meets the
circumference: And because DA is equal
to DC, the angle DAC is equal (1. 5.) to B
the angle ACD; but DAC is a right
angle, therefore ACD is a right angle,
and the angles DAC, ACD are therefore

equal to two right angles; which is impossible (1. 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.