Book III. join FA, AG: And because F is the centre of the circle ABC, E B A F с. G D whole FG is greater than FA, AG: But it is also less;a 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 A. Therefore, “ if two circles," &c. Q. E. D. a 20. 1. See N. 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 a 10. 11. 1. points B, D; join BD, and draw a GH bisecting BD at right angles: Therefore, because the points B, D are in the cir cumference of each of the circles, the straight line BD falls b 2. 3. within each of them: And their centres are in the straight e Cor. 1. 3. line GH which bisects BD at right angles: Therefore GH dll. 3. passes through the point of contact;d 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 Book III. touch the circle ABC in the points A, C, and join AC: Therefore, because the two points A, C are in the circumference of the circle ACK, K the straight line AC which joins them must fall within b the circle ACK: And b 2. 3. the circle ACK is without the circle ABC; and therefore the straight line A AC is also without the circle ABC; 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 than one point: B В 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 are equally distant from the centre. Take E the centre of the circle ABDC; from E draw EF, EG perpendiculars to AB, CD, and join AE and EC: Then because the straight line EF, passing through the centre, cuts the straight line AB, which does not C С pass through the centre, at right angles, it also bisects a it: Wherefore, AF is A a 3. 3. equal to FB, and AB double AF. For G the same reason, CD is double CG: And AB is equal to CD; therefore AF F is equal to CG; And because AE is D equal to EC, the square of AE is equal to the square of EC: But the squares B of AF, FE are equal to the square of b 47. I. 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 centre. Book III. 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 when the perpendiculars drawn to them from the centre are c 4. Def. 3. equal : c Therefore, AB, CD are equally distant from the 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, he demonstrated, that AB is double AF, and CD double 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: And AB is double AF, and CD double CG: wherefore AB is equal to CD. Therefore, “ equal straight lines,” &c. Q. E. D. PROP. XV. THEOR. See N. 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 AB F IK H D And, because BC is nearer to the centre than FG, EH is a 20. 1. less than EK: But, as was demonstrated in the preceding, Book III. BC is double BH, and FG double FK, and the squares of EH, HB are equal to the squares of EK, KF, of wbich the b 5. Def. 3. 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 FK: 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. PROP. XVI. THEOR. The straight line drawn at right angles to the dia. See N. meter 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 of the diameter, 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 where it meets the circumference: And because DA is equal to DC, the angle DAC is equala to the angle B A . a 5. 1. ACD; but DAC is a right angle, therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles; which is impossible. Therefore, the straight line drawn from b 17. 1. С D Book III. 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. 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 FÅ be between them, and c 12. 1. from the point D draw DG perpendicular to FA, and let it meet the circumference in H: and because AGD is a right d 19. 1. angle, and DAG less than a right angle; DA is greaterd than DG: But DA is equal to DH; E the less than the greater, which is с H B A or, which amounts to the same D 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 passes between that straight line and the perpendicular A E. " 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 e 2. 3. fall within it. " Also it is evident, that there can be but one straight line which touches the circle in the same point.” e 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; |