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, shall pass through the point A. For, if not, let it fall otherwise, if possible, as FGDH, and draw AF, AG. Then, because two sides of a triangle are together greater than the third side, there +20. 1. +5 Ax. 112 Def. 1 B fore FG, GA are greater than FA: but FA is equalt +12 Def. 1. to FH; therefore FG, GA are greater than FH: take away the common part FG; therefore the remainder AG is greater than the remainder GH; but AG is equal to GD; therefore GD is greater than GH, which is impossible, even should D, H coincide. 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 two circles, &c. Q. E. D. PROP. XII. THEOR. 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 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. B 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; wherefore the whole FG is not 12 Ax. F CD G *20. 1. less than FA, AG, even should C, D coincide: but it is less; which is impossible: therefore the straight line which joins the points F, G, cannot pass otherwise than through the point of contact A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D. PROP. XIII. THEOR. One circle cannot touch another in more points than one, whether it touches it on the inside or on the 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 B, D; and draw* GH bisecting BD at right angles: therefore, because the points *10. 11. 1. #2. 3. *Cor. 1.3. B, D are in the circumference of each of the circles, the straight line BD falls within each of them; therefore their centres are in the straight line GH which bisects BD at right angles; therefore GH passes through the points of contact: but it does not pass through them, 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. #11. 3. 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 A, C: therefore, because the two points A, C are in the circumference of the circle ACK, the straight line 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, &c. Q. E. D. PROP. 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 shall be equally distant from the centre. +1. 3. +12. 1. #3. 3. +Hyp. +7 Ax. Taket E the centre of the circle ABDC, and from pass through the centre, at right +12 Def. 1. EA is equal to EC, the square of EA is equal to the square of EC: but the squares of AF, FE are equal to the square of EA, because the angle AFE #47. 1. +1 Ax. +3 Ax. 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 to the squares of CG, GE: but the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of EF is equal 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 perpen#3 Def. 3. diculars drawn to them from the centre are equal: therefore AB, CD are equally distant from the centre. 13 Def. S. Next, let the straight lines AB, CD be equally distant from the centre, that is,† let FE be equal to EG; AB shall be 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: but 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 to the remaining square of CG; and the straight line AF is therefore equal to CG: but AB was shown to be double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. +Hyp. +3 Ax. +6 Ax. Q. E. D. PROP. 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 shall be greater than any straight line BC, which is not a diameter, and BC shall be greater than FG. AB F H +12. 1. From the centre draw† EH, EK perpendiculars to BC, FG, and join EB, EC, EF: and because +12 Def. 1. AE is equal† to EB, and ED to EC, therefore AD is equal to EB, EC: but EB, EC are greater* than BC; wherefore also AD is greater than BC. +2 Ax. *20. 1. +Hyp. And, because BC is nearert to the centre than FG, EH is less than EK: butt BC is double of BH, and FG double of FK, and, as was demonstrated in the preceding, the squares of EH, HB are equal to the squares of EK, KF: but 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 shall be nearer to the centre than FG, that is,† the same construction +4 Def. 3. being made, EH shall be 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 BH is greater than FK. Therefore the square of EH is less than the square of EK, and the straight line EII less than EK: and therefore BC is nearert to the centre than FG. Wherefore the diameter, &c. Q. E. D. +4 Def. 3. PROP. 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 from the extremity between that straight line and the circumference, so 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 a part of it fall, if possible, within |