wherefore the whole FG is greater than FA, *20. 1. AG: but it is also* 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 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, *10.11.1. and draw* GH bisecting BD at right angles: therefore, because the points B, D are in the *2.3. 3. * circumferences of each of the circles, the straight line BD falls within each of them; therefore *Cor. 1. their centres are * in the straight line GH which bisects BD at right angles; therefore GH *11. S. passes through the point of contact: 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: join AC: therefore, because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them falls within* the circle ACK: but the circle ACK is without the circle ABC; there fore the straight line AC is without this last circle: but because the points A, C are in the circumference of the circle ABC, the straight line AC must be within* the same circle, which is absurd : therefore one circle cannot touch another on the outside in more B K 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. * 2.3. + Hyp. *2.3. 3.3. Taket E the centre of the circle ABDC, and +1.3. from itt draw EF, EG perpendiculars to AB, † 12. 1. CD, and join EA, EC. Then, because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, A at right angles, it also bisects* it: therefore AF is equal to FB, and AB double of AF: for the same reason CD is double of CG: but B AB is equal to CD: therefore AF is equal to CG. And because AE is equal † +7 Ax. to EC, the square of AE is equal to the square +15 Def. 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; there +Hyp. +1 Ax. fore 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 +3 Ax. 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 perpendiculars *4Def.3. drawn to them from the centre are * equal · therefore AB, CD are equally distant from the centre. Next, let the straight lines AB, CD be equally +4Def.3. distant from the centre, that is,t 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 +Hyp. EG, because FE is equal to EG; therefore the +3 Ax. 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; where46 Ax. fore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D. +12. 1. 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. From the centre draw † EH, EK H 1. perpendiculars to BC, FG, and join EB, EC, EF: and because AE is equal to EB, and ED †15 Def. to EC, therefore AD is equal to EB, EC: but †2 Ax. EB, EC are greater than BC; wherefore also *20. 1. AD is greater than BC. And, because BC is nearer+ to the centre than + Hyp. FG, EH is less than EK: but, as was demon- *5Det.3. strated in the preceding, BC is double of BH, and FG double of FK, and 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 +5Def.3. same construction 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 EH less than EK: and therefore BC is nearer† to the centre +5Def.3. than FG. Wherefore the diameter, &c. Q.E.D. 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; 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 1. 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 +15 Def. because DA is equal† to DC, the angle DAC is equal to the angle ACD: but DAC is Hyp. aright angle; therefore ACD * 5. 1. B A is a right angle; and therefore the angles DAC, *17. 1. ACD are equal to two right angles; which is * impossible therefore the straight line drawn * See fig. 2. from A at right angles to BA does not fall within the circle. In the same manner it may be demonstrated, that it does not fall upon the circumference; therefore it must fall without the circle, as AE.* Also, between the straight line AE and the circumference no straight line can be drawn from the point A which does not cut the circle. For, if possible, let AF be between them: from the *12. 1. point D draw* DG perpendicular to AF, and let it meet the circumference in H. And because AGD is a right angle, and DAG less than a right angle, DA is greater than DG: but DA +15 Def. is equal to DH; therefore * 17. 1. 1. 19. 1. DH is greater than DG, the less than the greater, which FK amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point A, or however small an angle it makes with AE, the circumference must pass between that straight line and the perpendicular AE. "And this is all that is to be understood, |