side 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. THEOREM. 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. For the same reason, F B Take E the centre of the circle ABDC, and from it (I. 12.) draw EF, EG, perpendiculars to AB, 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, at right angles, it also bisects it (III. 3.), wherefore 1. AF is equal to FB, and AB double of AF. therefore E 2. CD is double of CG: but AB is equal (Hyp.) to CD; therefore (Ax. 7.) 4. The square of AE is equal to the square of EC; but (I. 47.) 5. The squares of AF, FE, are equal to the square of AE, because the angle AFE is a right angle; and, for the like reason, 6. The squares of EG, GC, are equal to the square of EC; 7. The squares of AF, FE, are equal 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 8. The remaining square of FE is equal to the remaining square of EG, and therefore 9. The straight line EF is 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 10. AB, CD, are equally distant from the centre E. 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 1. AB is double of AF, and CD double of CG, and that 2. 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 (Hyp.) to EG; therefore 3. The remaining square of AF is equal to the remaining square of CG; and therefore 4. The straight line AF is equal to CG: but AB is double of AF, and CD double of CG; wherefore 5. AB is equal to CD. Therefore, equal straight lines, &c. Q.E.D. 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. 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. G E H D From the centre E draw EH, EK, perpendiculars to BC, FG, and join EB, EC, EF. And because AE is equal to EB, and ED to EC, AD is equal to EB, EC: 1. but EB, EC, are greater (I. 20.) than BC; wherefore also 2. AD is greater than BC. And because BC is nearer (Hyp.) to the centre than FG, (III. Def. 5.) 1. EH is less than EK; but, as was demonstrated in the preceding proposition, and 3. The squares of EH, BH, are equal to the squares of EK, KF, of which the square of EH is less than the square of EK, because EH is less than EK; therefore 4. The square of BH is greater than the square of FK, and the straight line BH greater than FK; and therefore 5. 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 (III. Def. 5.) than EK. Because BC is greater than FG, likewise 1. BH is greater than KF: and The squares of BH, HE, are equal to the squares of of which the square of BH is greater than the square of FK, because BH is greater than FK; therefore 2. FK, KE, 3. The square of EH is less than the square of EK, and the straight line EH less than EK. . Wherefore, the diameter, &c. Q.E.D. PROP. 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 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 (I. Def. 15.) to DC, (I. 5.) 1. The angle DAC is equal the angle ACD; but DAC is (Hyp.) a right angle, therefore ACD is a right angle, and therefore 2. The angles DAC, ACD, are equal to two right angles; which is impossible (I. 17.): therefore 3. 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. 4. The straight line drawn from A at right angles to BẢ must fall without the circle, as AE. And 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 (I. 12.) DG perpendicular to FA, and let it meet the circumference in H. F E And because AGD is a right angle, and DAG less (I. 17.) than a right angle, (I. 19.) 1. DA is greater than DG: but DA is equal to DH; therefore the less than the greater, which is impossible: therefore 3. 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 straight line makes with the diameter at the point 4, or however small an angle it makes with AE, the circumference passes between the 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." 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. First, let A be a given point without the given circle BCD; it is required to draw a straight line from 4 which shall touch the circle. Find (III. 1.) the centre E of the circle, and join AE; and from the centre E, at the distance AE, describe the circle AFG; from the point D draw (I. 11.) FD at right angles to AE; and join EBF, AB: AB touches the circle BCD. Because E is the centre of the circles BCD, AFG; (I. Def. 15.) 1. EA is equal to EF, 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 (I. 4.) 2. 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 3. The angle EBA is equal to the angle EDF: but EDF is (Constr.) a right angle, wherefore 4. EBA is a right angle: 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 5. AB touches the circle; and it is drawn from the given point A. Which was to be done. 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; (III. 16. Cor.). 1. DF touches the circle. |