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 (1. 11.) 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 AD is equal (Ax.) to EM, MD:

But EM, MD are greater (20. 1.) 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: but the angle EMD is greater (9 Ax.) than the angle FMD; therefore the base ED is greater (24. 1.) than the base FD.

In like manner it may be shown that FD is greater than CD: therefore DA is the greatest: c and DE greater than DF, and DF than DC.

And because MK, KD are greater (20. 1.) than MD, and MK is equal (15 Def. 1.) to MG, the remainder DK is greater (5 Ax.) than the remainder GD; that is, GD is less than KD:

A

M

And because MK, DK are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD, MK, KD are less (21. 1.) than ML, LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL:

In like manner it may be shown, that 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 (23. 1.) the angle DMB equal to the angle DMK, 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; and the angle KMD is equal (Constr.) to the angle BMD; therefore the base DK is equal (4. 1.) to the base DB:

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, the nearer to the least equal to the more remote, which is impossible. If therefore any point, etc. Q. E. D.

PROPOSITION IX.

THEOR. 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 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, DG shall be the greatest line from it to the circumference, and DC greater (7. III.) 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: 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, etc.

PROPOSITION X.

Q. E. D.

THEOR. 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, and join KB, KG, KF :

And because within the circle DEF there is taken the point K, from which to the cir- E cumference DEF fall more than two equal straight lines KB, KG, KF, the point K is (9. III.) the centre of the circle DEF:

Λ

B

H

D

c

K

But K is also the centre (Constr.) of the circle ABC; therefore the same point is the centre of two circles that cut one another, which is impossible (5. III.). Therefore one circumference of a circle cannot cut another in more than two points. Q. E. D.

PROPOSITION XI.

THEOR. If one circle touch another 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 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 (20. 1.) than FA, that is, than FH, for FA is equal (15 Def. 1.) to FH; take away the common part FG; therefore the remainder AG is greater than the remainder GH:

H

D

F

E

B

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, etc.

Q. E. D.

PROPOSITION XII.

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

B

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.

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 (2 Ax.) to FC, DG; wherefore the whole FG is greater than FA, AG: but it is also less (20. 1.), 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, etc.

PROPOSITION XIII.

Q. E. D.

THEOR. 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 (11. 1.) 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 (2.111.) of them: and their centres are (J. III.) in the straight line GH which bisects BD at right angles therefore GH passes through the point of contact (11. III.);

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.

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 (2. 111.) the circle ACK:

And the circle ACK is without the circle ABC; and therefore the straight line AC is without (Hyp.) this last circle:

But because the points A, C are in the circumference B

of the circle ABC, the straight line AC must be within (2. III.) the same circle, which is absurd; therefore 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, etc. Q. E. D.

PROPOSITION 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.

Take (1. III.) E, the centre of the circle ABDC, and from it (12. 1.) 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 (3. III.) 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 (Hyp.) to CD; therefore AF is equal (7 Ax.) to CG:

F

E

And because AE is equal (15 Def. 1.) to EC, the square of AE is equal to the square of EC;

But the squares of AF, FE are equal (47. 1.) to 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 (1 Ax.) 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 (3 Ax.) 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 (4 Def. 111.); therefore AB, CD must be equally distant from the

centre.

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 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 (3 Ax.) to the remaining square of CG; and the straight line AF is therefore equal to CG;

And AB is double of AF, and CD double of CG: wherefore AB is equal (6 Ax.) to CD. Therefore equal straight lines, etc. Q. E. D.

PROPOSITION 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 (12. 1.) EH, EK, perpendiculars to BC, FG, and join EB, EC, EF;

And because AE is equal (15 Def. 1.) to EB, and ED to EC, AD is equal to EB, EC:

But EB, EC are greater (20. 1.) than BC; wherefore also AD is greater than BC.

And, because BC is nearer to the centre than FG, EH is less (5 Def. III.) than EK:

B

K

H

D

But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares EK, 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 BII 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, etc. Q. E. D.

PROPOSITION 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 possible, within the circle, as AC, and draw DC to the point C, where it meets the circumference:

And because DA is equal (15 Def. 1.) to DC, the

angle DAC is equal (5. 1.) to the angle ACD: but DAC is a right (Hyp.) angle; therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles, which is impossible (17. 1.); therefore the straight line drawn from A at right angles to BA does not fall within the circle.