PROP. 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. (References-Prop. I. 12, 20; 11. def. 5.) 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. CONSTRUCTION From the centre E draw EH, EK, perpendiculars to BC, FG, (1. 12) and join EB, EC, EF. DEMONSTRATION Then, because AE is equal to EB, and ED to EC, (1. def. 15) therefore AD is equal to EB, EC; (ax. 2) but EB, EC, are greater than BC; (1. 20) wherefore also AD is greater than BC. And because BC is nearer to the centre than FG, (hyp.) therefore EH is less than EK; (III. def. 5) 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 the straight line BH greater than 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, EH shall be less than EK. The same construction being made, because BC is greater than FG, therefore BH is greater than KF; and the squares of BH, HE, are equal to the squares of FK, KE, but 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 than FG. (III. def. 5.) Wherefore, the diameter, &c. Q. E. D. PROP. XVI.-THEOREM. The straight line which is 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, which does not 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. (References-Prop. 1. 5, 12, 17, 19; III. 2, def. 2.) 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. 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. DEMONSTRATION Because DA is equal to DC, (1. def. 15) the angle DAC is equal to the angle ACD; (1. 5) but DAC is a right angle, (hyp.) therefore ACD must be a right angle, and therefore the angles DAC, ACD, must be equal to two right angles; which is impossible; (1. 17) therefore the straight line drawn from ▲ 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 the straight line from A at right angles to BA, 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 FA be between them, and from the point D draw DG perpendicular to FA, (1. 12) and let it meet the circumference in H. DEMONSTRATION Because AGD is assumed to be a right angle, and DAG less than a right angle; (1. 17) therefore DA must be greater than DG; (1. 19) but DA is equal to DH; (1. def. 15) therefore DH must be greater than DG, 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 shall pass between that straight line and the perpendicular AE. 'And this is all that is to be understood, when in the Greek text, and in 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; (III. 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.' PROP. XVII.-PROBLEM. To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. (References-Prop. I. 4, 11; III. 1, 16, cor.) 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, (III. 1) and join AE; and from the centre E, at the distance AE, describe the circle AFG; from the point D draw FD at right angles to AE; (L. 11) and join EBF, AB. Then AB shall touch the circle BCD. DEMONSTRATION Because E is the centre of the circles BCD, AFG; therefore EA is equal to EF, and ED to EB; (1. def. 15) hence, in the two triangles AEB, FED, because the two sides AE, EB, are equal to the two FE, ED, each to each; and they contain the angle at E common to both triangles; therefore, the base DF is equal to the base AB, 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; (1. 4) but EDF is a right angle, (constr.) wherefore EBA is a right angle; (ax. 1) and EB is drawn from the centre; but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle;' (III. 16, cor.) therefore AB touches the circle, and it is drawn from the given point A. But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE; then DF touches the circle. (III. 16, cor.) Q. E. F. PROP. XVIII.—THEOREM. If a straight line touch a circle, the straight line drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle. (References-Prop. 1. 17, 19; ш. 1.) Let the straight line DE touch the circle ABC in the point C; take the centre F, (III. 1) and draw the straight line FC. Then FC shall be perpendicular to DE. For, if it be not, from the point F draw FBG perpendicular to DE. |