Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC, be drawn to the circumference, whereof DA passes through the centre. Of those which fall upon the concave part of the circumference AEFC, the greatest is AD which passes through the centre; and the nearer to it is always greater than the more remote, viz. DE than DF, and DF than DC: but of those which fall upon the convex circumference HLKG, the least is DG between the point D and the diameter AG; and the nearer to it is always less than the more remote, viz. DK than DL, and DL than DH.

Take (III. 1.) M the centre of the circle ABC, and join ME, MF, MC, MK, ML, MH. And because AM is equal to ME, and MD to each, therefore (Ax. 2.)

1. AD is equal to EM, MD;

but EM, MD, are greater (I. 20.) than ED; therefore also

2. AD is greater than ED.

Again, because ME is equal to MF, and MD common to the triangles EMD, FMD; EM, MD, are equal to FM, MD, each to each, but the angle EMD is greater than the angle FMD; therefore (I. 24.)

3.

The base ED is greater than the base FD.

In like manner it may be shown that

4. FD is greater than CD:

therefore DA is the greatest, and DE greater than DF, and DF than DC.

And because MK, KD, are greater (I. 20.) than MD, and MK is equal (I. Def. 15.) to MG, the remainder KD is greater (Ax. 5.) than the remainder GD; that is,

1. GD is less than KD;

and because MK, DK, are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD, therefore (I. 21.) · 2. MK, KD, are less than ML, LD,

whereof MK is equal to ML; therefore

3. The remainder DK is less than the remainder DL.

In like manner it may be shown, that

4. DL is less than DH:

therefore DG is the least, and DK less than DL, and DL than DH.

Also, there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least. At the point M, in the straight line MD, make (I. 23.) the angle DMB equal to the angle DMK, and join DB. And because MK is equal to MB, and MD common to the triangle KMD, BMD, the two sides KM, MD, are equal to the two BM, MD, each to each; and the angle KMD is equal (Constr.) to the angle BMD; therefore (I. 4.)

but,

1.

2.

The base DK is equal to the base DB:

Besides DB there can be no straight line drawn from

D to the circumference equal to DK;

for, if there can, let it be DN; and because DK is equal to DN, and also to DB; therefore

3. DB is equal to DN,

that is, a line nearer to the least is equal to the more remote, which is impossible. If, therefore, any point, &c. Q.E.D.

PROP. IX.-THEOREM.

If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle.

Let the point D be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC, the point D is the centre of the circle.

For, if not, let E be the centre, join DE and produce it to the circumference in F, G; then FG is a diameter (I. Def. 17.) of the circle ABC: and because in FG, the diameter of the circle ABC, there is taken the point D, which is not the centre, (III. 7.)

1. DG is the greatest line from the point D to the circumference, and DC is greater than DB, and DB than DA:

but they are likewise equal (Hyp.), which is impossible: therefore E is not the centre of the circle ABC,

2.

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

2. AG, GF, are greater than FH;

take away 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 4; 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

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:

also, because G is the centre of the circle ADE, 2. AG is equal to GD:

therefore FA, AG, are equal to FC, DG; wherefore 3. The whole FG is greater than FA, AG;

but also (I. 20.)

which is impossible: therefore the straight line which joins the points F, G, shall not pass otherwise than through the point of contact 4; 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 (IIÍ. 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, Č.

K

Join AC. 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.)

3. The straight line AD must be within the circle ABC, which is absurd: therefore one circle cannot touch another on the out