therefore the straight line which joins the points F, G, cannot pass that is, FG must pass through the point A. PROPOSITION XIII. THEOREM. 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 GH bisecting BD at right angles. (1. 11.) Because the points B, D are in the circumferences of each of the circles, therefore the straight line BD falls within each of them; (III. 2.) therefore their centres are in the straight line GH which bisects BD at right angles; (III. Cor. 1.) therefore GH passes through the point of contact: (III. 11.) because the points B, D are without the straight line GH; therefore one circle cannot touch another on the inside in more 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; B join AC. A K Because the two points A, C are in the circumference of the circle ACK, therefore the straight line AC which joins them, falls within the circle ACK: (III. 2.) but the circle ACK is without the circle ABC; (hyp.) therefore 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, (111. 2.) which is absurd; therefore one circle cannot touch another on the outside in more than one point: and it has been shewn, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q.E.D. PROPOSITION XIV. THEOREM. Equal straight lines in a circle are equally distant from the centre; and conversely, 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. Then AB and CD shall be equally distant from the centre. Take E the centre of the circle ABDC, (111. 1.) from E draw EF, EG perpendiculars to AB, CD, (1. 12.) 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, at right angles, it also bisects it: (III. 3.) therefore AF is equal to FB, and AB double of AF. And because AE is equal to EC, (1. def. 15.) the but the squares of AF, FE are equal to the square of AE, because the angle AFE is a right angle; (1.47.) 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: (ax. 1.) 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, (ax. 3.) 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: (III. def. 4.) therefore AB, CD are equally distant from the centre. Next, let the straight lines AB, CD be equally distant from the centre, (III. def. 4.) that is, let FE be equal to EG; then 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; (hyp.) therefore the remaining square of AF is equal to the remaining square of CG: (ax. 3.) and the straight line AF is therefore equal to CG: but AB was shewn to be double of AF, and ĈD double of CG; wherefore AB is equal to CD. (ax. 6.) Therefore equal straight lines, &c. Q.E.D. PROPOSITION XV. THEOREM. 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. Then AD shall be greater than any straight line BC, which is not a diameter, and BC shall be greater than FG. F A B D From the centre E draw EH, EK perpendiculars to BC, FG, (1. 12.) and join EB, EC, EF. And because AE is equal to EB, and ED to EC, (1. def. 15.) but EB, EC are greater than BC; (1. 20.) And, because BC is nearer to the centre than FG, (hyp.) but, as was demonstrated in the preceding proposition, 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 therefore BC is greater than FG. Next, let BC be greater than FG; then BC shall be nearer to the centre than FG, that is, the same construction being made, EH shall be less than EK. (111. def. 5.) 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, therefore the square of EH is less than the square of EK, and therefore BC is nearer to the centre than FG. (111. def. 5.) Wherefore the diameter, &c. Q.E.D. PROPOSITION XVI. THEOREM. 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 D, and the diameter AB. Then the straight line drawn at right angles to AB from its extremity A, shall fall without the circle. A 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 to DC, (1. def. 15.) the angle DAC is equal to the angle ACD: (1. 5.) and therefore the angles DAC, ACD are equal to two right angles; which is impossible: (1. 17.) therefore the straight line drawn 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 point D draw DG perpendicular to AF, (1. 12.) And because AGD is a right angle, and DAG less than a right angle, (1. 17.) therefore DA is greater than DG: (1. 19.) the less than the greater, which is impossible: therefore no straight line can be drawn from the point A, between AE and the circumference, which does not cut the circle: or, which 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, when, in the Greek text, and translations from it, the angle of the semicircle is said to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal angle." Q.E.D. COR. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle; (111. def. 2.) and that it touches it only in one point, because, if it did meet the circle in two, it would fall within it. (III. 2.) "Also, it is evident, that there can be but one straight line which touches the circle in the same point." PROPOSITION XVII. PROBLEM. To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. First, let A be a given point without the given circle BCD; it is required to draw a straight line from A which shall touch the circle. Find the centre E of the circle, (111. 1.) and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw DF at right angles to EA, (1. 11.) and join EBF, AB. Then AB shall touch the circle BCD in the point B. Because E is the centre of the circles BCD, AFG, therefore EA is equal to EF, (1. def. 15.) and ED to EB; therefore the two sides AE, EB, are equal to the two FE, ED, each to each; and they contain the angle at E common to the two triangles AEB, FED; therefore the base DF is equal to the base AB, (1. 4.) and the triangle FED to the triangle AEB, and the other angles to the other angles: therefore the angle EBA is equal to the angle EDF: but EDF is a right angle, (constr.) wherefore EBA is a right angle: (ax. 1.) and EB is drawn from the centre: |