bisects another 4C which does not pass through the centre, it shall cut it (III. 3.) at right angles; wherefore

1. FEA is a right angle.

Again, because the straight line FE bisects the straight line BD which does not pass through the centre, it shall cut it at right angles (III. 3.); wherefore

· 2. FEB is a right angle:

and FEA was shown to be a right angle; therefore 3. FEA is equal to the angle FEB,

the less to the greater, which is impossible: therefore 4. AC, BD, do not bisect one another.

Wherefore, if in a circle, &c.

Q.E.D.

PROP. V.-THEOREM.

If two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG, cut one another in the points B, C; they have not the same centre

A

G

For, if it be possible, let E be their centre: join EC, and draw any straight line EFG meeting them in F and G. And because E is the centre of the circle ABC, (I. Def. 15.)

again, because E is the centre of the circle CDG,

but CE was shown to be equal to EF; therefore

3. EF is equal to EG,

the less to the greater, which is impossible; therefore

4. E is not the centre of the circles ABC, CDG.

Wherefore, if two circles, &c.

Q.E.D.

PROP. VI.-THEOREM.

If two circles touch one another internally, they shall not have the same centre.

Let the two circles ABC, CDE, touch one another internally in the point C; they have not the same centre.

For, if they can, let the centre he F; join FC, and draw any straight line FEB meeting them in E and B; and because F is the centre of the circle ABC, (I. Def. 15.)

but CF was shown to be equal to FB; therefore (Ax. 1.) 3. FE is equal to FB,

the less to the greater, which is impossible; wherefore

4. Fis not the centre of the circles ABC, CDE.

Therefore, if two circles, &c. Q.E.D.

PROP. VII.-THEOREM.

If any point be taken in the diameter of a circle, which is not the centre, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line which passes through the centre is always greater than one more remote: and from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line.

Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre: let the centre be E: of all the straight lines FB, FC, FG, &c., that can be drawn from F to the circumference, FA is the greatest, and FD, the other part of the diameter AD, is the least and of the others, FB is greater than FC, and FC than FG.

A
B

K

Join BE, CE, GE; and because two sides of a triangle are greater than the third, (I. 20.)

but AE is equal (I. Def. 15.) to EB; therefore

2. AE, EF, that is, AF, is greater than BF.

Again, because BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides BE, EF, are equal to the two CE, EF, each to each; but the angle BEF is greater than the angle CEF; therefore (I. 24.)

3. The base BF is greater than the base FC:

for the same reason,

4. CF is greater than GF.

Again, because GF, FE, are greater (I. 20.) than EG, and EG is equal to ED; therefore

6. The remainder GF is greater than the remainder FD. Therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF.

Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD. At the point E, in the straight line EF, make (I. 23.) the angle FEH equal to the angle GEF, and join FH. Then because GE is equal (I. Def. 15.) to EH, and EF common to the two triangles GEF, HEF, the two sides GE, EF, are equal to the two HE, EF, each to each; and the angle GEF is equal (Constr.) to the angle HEF; therefore (I. 4.)

but

1.

The base FG is equal to the base FH:

2. Besides FH, no other straight line can be drawn from F to the circumference equal to FG ;

for, if there can, let it be FK; and because FK is equal to FG, and FG to GH,

3. FK is equal to FH;

that is, a line nearer to that which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if any point be taken, &c. Q.E.D.

If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre; and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote: but of those which fall upon the convex circumference, the least is that between the point without the circle and the diameter; and of the rest, that which is nearer to the least is always less than one more remote: and only two equal straight lines can be drawn from the same point to the circumference, one upon each side of the least line.

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.

D

BN

M

A

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, add 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 2. E is not the centre of the circle ABC.