In like manner it may be demonstrated that no other point but D is the centre; therefore

3. D is the centre of the circle ABC.

Wherefore if a point be taken, &c. Q.E.D.

PROP. X.-THEOREM.

One circumference of a circle cannot cut another in more than two points. If it be possible, let the circumference FAB cut the circumference DEF in more than two points, viz. in B, G, F.

Take the centre K (III. 3.) 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, (III. 9.)

1. The point K is the centre of the circle DEF;

but K is also the centre of the circle ABC; therefore

2. The point K is 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.-THEOREM.

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

Let the two circles ABC, ADE, touch each other internally in the point 4, 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 4.

For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG.

And because two sides of a triangle are together greater than the third side, (I. 20.)

1. AG, GF, are greater than FA;

but FA is equal (I. Def. 15.) to FH, both being from the same centre ; therefore

take away

2. AG, GF, are greater than FH;

the common part FG; therefore

3. The remainder AG is greater than the remainder GH;

but AG is equal (I. Def. 15.) to GD; therefore

4. 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 4, that is,

5. FG being produced must pass through the point A.

Therefore, if two circles, &c.

Q.E.D.

PROP. XII.-THEOREM.

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 ; 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 4.

B

E

F C

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,

1. AF is equal to FC:

2.

also, because G is the centre of the circle ADE,
AG is equal to GD:
therefore FA, AG, are equal to FC, DG; wherefore
3. The whole FG is greater than FA, AG;

which is impossible: therefore the straight line which joins the points F, G, shall not pass otherwise than through the point of contact ; that is, 5. FG must pass through the point A.

Therefore, if two circles, &c. Q.E.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 (I. 10, 11.) GH bisecting BD at right angles. And because the points B, D, are in the circumference of each of the circles, (III. 2.)

1. The straight line BD falls within each of the circles EBF, ABC;

and their centres are (III. 1. Cor.) in the straight line GH which bisects BD at right angles: therefore (III. 11.)

but

2.

GH passes through the point of contact;

3. GH does not pass through the point of contact,

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.

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.

Join AC.

K

Then, because the two points A, C, are in the circumference of the circle ACK, (III. 2.)

1.

The straight line AC falls within the circle ACK: but the circle ACK is without (Hyp.) the circle ABC; therefore 2. The straight line AC is without the circle ABC;

but, because the points A, C, are in the circumference of the circle ABC, (III. 2.)

C

3. The straight line AD must be within the circle ABC,

which is absurd: therefore one circle cannot touch another on the out

side 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.-THEOREM.

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.

F

B

Take E the centre of the circle ABDC, and from it (I. 12.) draw EF, EG, perpendiculars to AB, CD, and join EA, EC. 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 it (III. 3.), wherefore

1. AF is equal to FB, and AB double of AF.

For the same reason,

2. CD is double of CG:

but AB is equal (Hyp.) to CD; therefore (Ax. 7.) AF is equal to CG.

3.

And because AE is equal (I. Def. 15.) to EC,

4.

but (I. 47.)

The square of AE is equal to the square of EC;

5. The squares of AF, FE, are equal to the square of AE, because the angle AFE is a right angle; and, for the like reason, 6. The squares of EG, GC, are equal to the square of EC;

therefore

7. The squares of AF, FE, are equal to the squares of OG, GE,

of which the square of AF is equal to the square of CG, because AF is equal to CG; therefore

1

8. The remaining square of FE is equal to the remaining square of EG,

and therefore

9. The straight line EF is 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

10. AB, CD, are equally distant from the centre E.

Next, if the straight lines AB, 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

and that

1. AB is double of AF, and CD double of CG,

2.

EG, GC;

The squares of EF, FA, are equal to the squares of

of which the square of FE is equal to the square of EG, because FE is equal (Hyp.) to EG; therefore

3. The remaining square of AF is equal to the remaining square of CG;

and therefore

4. The straight line AF is equal to CG:

but AB is double of AF, and CD double of CG; wherefore 5. AB is equal to CD.

Therefore, equal straight lines, &c. Q.E.D.

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.

B

G

D

From the centre E draw EH, EK, perpendiculars to BC, FG, and join EB, EC, EF. And because AE is equal to EB, and ED to EC,

1. AD is equal to EB, EC:

but EB, EC, are greater (I. 20.) than BC; wherefore also

2. AD is greater than BC.

And because BC is nearer (Hyp.) to the centre than FG, (III. Def. 5.)

1. EH is less than EK;