therefore FE is equal to FB, (ax. 1.)

the less equal to the greater, which is impossible: therefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c.

Q. E.D.

PROPOSITION 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 equal straight lines to the circumference, one upon each side of the diameter.

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.

Then, of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference,

FA, that in which the centre is, shall be the greatest, and FD, the other part of the diameter AD, shall be the least:

and of the others, FB, the nearer to FA, shall be greater than FC, the more remote, and FC greater than FG.

Join BE, CE, GE.

And because two sides of a triangle are greater than the third, (1. 20.) therefore BE, EF are greater than BF: but AE is equal to BE; (1. def. 15.) therefore 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; (ax. 9.) therefore the base BF is greater than the base CF. (1. 24.) For the same reason CF is greater than GF. Again, because GF, FE are greater than EG, (1. 20.) and EG is equal to ED;

therefore GF, FE are greater than ED:

take away the common part FE,

and the remainder GF is greater than the remainder FD. (ax. 5.) 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 diameter.

At the point E, in the straight line EF, make the angle FEH equal to the angle FEG, (1. 23.) and join FH.

Then, because GE is equal to EH, (1. def. 15.)

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 to the angle HEF; (constr.) therefore, the base FG is equal to the base FH: (1. 4.)

but, 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 FH,
therefore FK is equal to FH; (ax. 1.)

that is, a line nearer to that which passes through the centre, is
equal to one which is more remote;

which has been proved to be 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 line which passes through the centre.

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 shall be DA, which passes through the centre ; and any line nearer to it shall be greater than one more remote, viz. DE, shall be greater than DF, and DF greater than DC: but of those which fall upon the convex circumference HLKG, the least shall be DG between the point D and the diameter AG; and any line nearer to it shall be less than one more remote, viz. DK less than DL, and DL less than DH.

Take M the centre of the circle ABC, (111. 1.)
and join ME, MF, MC, MK, ML, MH.
And because AM is equal to ME,

add MD to each of these equals,

therefore AD is equal to EM, MD:

(ax. 2.)

(1. 20.)

but EM, MD are greater than ED;
therefore also 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; (ax. 9.)
therefore the base ED is greater than the base FD. (1. 24.)
In like manner it may be shewn that FD is greater than CD.
Therefore, DA is the greatest;

and DE greater than DF, and DF greater than DC.
And, because MK, KD are greater than MD, (1. 20.)
and MK is equal to MG, (1. def. 15.)

the remainder KD is greater than the remainder GD, (ax. 5.)
that is, GD is less than KD:

and because MLD is a triangle, and from the points M, D, the extremities of its side MD, the straight lines MK, DK are drawn to the point K within the triangle,

therefore MK, KD are less than ML, LD: (1. 21.)

but MK is equal to ML; (1. def. 15.)

therefore, the remainder DK is less than the remainder DL. (ax. 5.) In like manner it may be shewn, that DL is less than DH. Therefore, DG is the least, and DK less than DL, and DL less 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 line through the

centre.

At the point M, in the straight line MD,

make the angle DMB equal to the angle DMK, (1. 23.) and join DB. And because MK is equal to MB, and MD common to the triangles KMD, BMD,

the two sides KM, MD are equal to the two BM, MD, each to each; and the angle KMD is equal to the angle BMD; (constr.) therefore the base DK is equal to the base DB: (1.4.)

but, 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 DB is equal to DN;

that is, a line nearer to the least is equal to one more remote, which has been proved to be impossible.

If, therefore, any point, &c. Q.E.D.

PROPOSITION 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.

Then the point D shall be 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 of the circle ABC: (1. def. 17.) and because in FG, the diameter of the circle ABC, there is taken the point D, which is not the centre,

therefore DG is the greatest line from it to the circumference, and DC is greater than DB, and DB greater than DA: (111. 7.) but they are likewise equal, (hyp.) which is impossible: therefore E is not the centre of the circle ABC.

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

Wherefore, if a point be taken, &c.

Q.E. D.

PROPOSITION 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 of the circle ABC, (111. 3.) and join KB, KG, KF. Then because K is the centre of the circle ABC,

therefore KB, KG, KF are all equal to each other: (1. def. 15.) 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,

therefore the point K is the centre of the circle DEF: (111.9.) but K is also the centre of the circle ABC; (constr.)

therefore the same point 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.

PROPOSITION XI. THEOREM.

If one circle touch another internally in any point, the straight line which joins their centres being produced shall pass through that point of

contact.

Let the circle ADE touch the circle ABC internally in the point A;

and let F be the centre of the circle ABC, and G the centre of the circle ADE;

then the straight line which joins the centres F, G, being produced, shall pass through the point A.

For, if FG produced do not pass through the point A, let it fall otherwise, if possible, as FGDH, and join AF, AG. Then, because two sides of a triangle are together greater than the third side, (1. 20.)

take

therefore FG, GA are greater than FA:

but FA is equal to FH; (1. def. 15.) therefore FG, GA are greater than FH:

away from these unequals the common part FG;

therefore the remainder AG is greater than the remainder GH; (ax. 5.) but AG is equal to GD; (1. def. 15.) therefore GD is greater than GH,

the less than the greater, which is impossible.

Therefore the straight line which joins the points F, G, being produced, cannot fall otherwise than upon the point A,

that is, it must pass through it.

Therefore, if one circle, &c. Q. E.D.

PROPOSITION XII. THEOREM.

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

Let the two circles ABC, ADE, touch each other externally in the point Д;

and let F be the centre of the circle ABC, and G the centre of ADE. Then 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,
FA is equal to FC:

also, because G is the centre 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; (1. 20.) which is impossible: