centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects (II. 3.) it: Wherefore A F is equal to FB, and AB double of a F. For the same F reason, CD is double of CG: And E square of EC: But the squares of AF, FE are equal (1. 47.) 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: Therefore 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 the remaining square of FE 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 perpendiculars drawn to them from the centre are equal (III. Def. 4.): Therefore AB, CD are equally distant from the centre. 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 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; of which 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 is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D. 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. From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; and because AE is F A B D H equal to E B, and ED to EC: AD is equal to EB, EC: but EB, EC, are greater (1. 20.) than BC; wherefore also AD is greater than BC. And because BC is nearer to the centre than FG, EH is less (III. Def. 5.) than EK: But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and 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 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 is nearer to the centre than FG, that is, the same construction being made, EH is 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. Wherefore the diameter, &c. Q. E. D. 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 to DC, the angle DAC is equal (1. 5.) to the angle ACD; but DAC is a right angle, therefore B ACD is a right angle, and the angles DAC, ACD are therefore 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 A E. And between the straight line AE and the cir F cumference 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: And because A GD is a right angle, and DA G less (1. 17.) than a right angle: DA is greater (1. 19.) than DG: But DA is equal to DH; therefore DH is greater than DG, the less than the greater, which is impossible: Therefore no straight line B can be drawn from the point A between AE and the circumference, which does not cut the circle, or, which G A 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 A E, 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; 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. PROB. 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 (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 (1. 11.) FD at right angles to AE; and join EBF, A B. touches the circle BC D. AB Because E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED; therefore the base DF is equal to the base AB, and the triangle FED to the G D E B triangle AE B, and the other angles to the other angles (1. 4.): Therefore the angle EBA is equal to the angle EDF: But EDF is a right angle, wherefore 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 (II. 16. Cor.): Therefore A B 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 |