more than one point: For, if it be poffible, let the circle ACK Book III. touch the circle ABC in the points A, C, and join AC: There fore, because the two points A, C are in K b 2. 3. on the infide in more points than one: Therefore, one circle, &c. Q. E. D. QUAL ftraight lines in a circle are equally diftant from the centre; and those which are equally diftant from the centre, are equal to one another. Let the ftraight lines AB, CD, in the circle ABDC, be equal to one another; they are equally distant from the centre. a C Take E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD: Then, because the straight line EF, paffing through the centre, cuts the ftraight line AB, which does not país through the centre, at right angles, it also bifects it: Wherefore AF is equal to FB, and AB double of AF. For the fame reafon, CD is double of CG: And AB is equal to CD; therefore AF is equal to CG: And because AE is equal to EC, the fquare F of AE is equal to the fquare of EC: But the fquares of AF, FE are equal to the fquare of AE, because the B angle AFE is a right angle; and, for the like reafon, the fquares of a 3.3. E D b 47. E. EG, GC are equal to the fquare of EC: Therefore the fquares of AF, FE are equal to the fquares of CG, GE, of Book III. which the fquare of AF is equal to the fquare of CG, because AF is equal to CG; therefore the remaining fquare of FE is equal to the remaining (quare of EG, and the ftraight line FE is therefore equal to EG: But ftraight lines in a circle are faid to be equally diftant from the centre, when the perpendiculars c 4. Def. 3. drawn to them from the centre are equal: Therefore AB, CD are equally diftant from the centre. Next, if the ftraight lines AB, CD be equally diftant from the centre, that is, if FE be equal to EG; AB is equal to CD: For, the fame conftruction being made, it may, as before, be demonftrated, that AB is double of AF, and CD double of CG, and that the fquares of EF, FA are equal to the fquares of EG, GC; of which the fquare of FE is equal to the fquare of EG, because FE is equal to EG; therefore the remaining fquare of AF is equal to the remaining fquare of CG; and the straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal ftraight lines, &c. Q. E. D. See N. 20. I. TH HE diameter is the greateft ftraight 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 de a circle, of which the diameter is AD, and centre E; and let BC be nearer to the centre than FG; AD is greater than any ftraight 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, LF; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC.: But EB, EC, are greater than BC; wherefore, alfo AD is greater than BC. a And, becaufe BC is nearer to the centre than FG, EH EH is less than EK: But, as was demonftrated in the pre- Book III. ceding, BC is double of BH, and FG double of FK, and the fquares of EH, HB are equal to the fquares of EK, KF, of s. Def. 3. which the fquare of EH is lefs than the fquare of EK, because EH is less than EK; therefore the fquare of BH is greater than the fquare of FK, and the ftraight 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 fame conftruction being made, EH is lefs than EK: Because BC is greater than FG, BH likewife is greater than FK: And the fquares of BH, HE are equal to the fquares of FK, KE, of which the fquare of BH is greater than the fquare of FK, because BH is greater than FK; there. fore the fquare of EH is lefs than the square of EK, and the ftraight line EH less than EK. Wherefore the diameter, &c. Q. E. D. THE HE ftraight 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, fo as not to cut the circle; or, which is the fame thing, no straight line can make fo great an acute angle with the diameter at its extremity, or fo fmall an angle with the ftraight 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, fhall fall without the circle. For, if it does not, let it fall, if poffible, within the circle as AC, and draw DC to the point C where therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles; which is impoffible b:b 17, I. F Therefore Book III. Therefore the ftraight line drawn from A at right angles to BA does not fall within the circle: In the fame manner, it may be. demonftrated that it does not fall upon the circumference; therefore it must fall within the circle, as AE. C 12. I. d 19. I. 2. 3. And between the ftraight line AE and the circumference no ftraight line can be drawn from the point A which does not cut the circle: For, if poffible, let FA be between them, and from the point D draw DG perpendicular to FA, and let it meet the circumference in H: And because AGD is a right angle, and DAG lefs than a right angle: DA is greater than DG: But DA is equal to DH; therefore DH is greater than DG, the less than the greater, which is impoffible: Therefore no ftraight line can be drawn from the point A between AE and the circumference, which does not cut the cir cle, or, which amounts to the fame B thing, however great an acute angle a ftraight line makes with the diameter at the point A, or however fmall an angle it makes with AE, F E C H A Ꭰ the circumference paffes between that ftraight line and the perpendicular AE. And this is all that is to be understood, when, in the Greek text and tranflations from it, the angle of the femicircle is faid to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal an gle.' COR. From this it is manifeft that the ftraight 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, becaufe, if it did meet the circle in two, it would fall within it. Also it is evident that there can be but one ftraight line which touches the circle in the fame point.' T O draw a ftraight 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 ftraight line from A which fhall touch Book III. the circle. Find the centre E of the circle, and join AE; and from a I. 3. the centre E, at the diftance EA, defcribe the circle AFG; b from the point D draw b DF at right angles to EA, and join II. I. EBF, AB; AB touches the circle BCD. Becaufe E is the centre of the circles BCD, AFG, EA is equal to EF: And ED to EB; therefore the two fides 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 bafe DF is equal to the bafe AB, and the tri angle FED to the triangle. AEB, and the other angles to the other angles: Therefore the c 4.1. 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 ftraight line drawn from the extremity of a diameter, at right angles to it, touches the circled: Therefore & Cor. 16 3. 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; DF touches the circled. Fa ftraight line touches a circle, the ftraight line drawn pendicular to the line touching the circle. Let the ftraight line DE touch the circle ABC in the point C, take the centre F, and draw the ftraight line FC; FC is perpendicular to DE. For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, GCF is an acute b 17. 1. angle; and to the greater angle the greateft fide is oppofite: c 19. 1. Therefore F 2 C |