Book III. join FA, AG: And because F is the center of the circle ABC, a 20. I. AF is equal to FC: Alfo Ε A F G Therefore the ftraight line which joins the points F, G fhall not pafs otherwife than through the point of contact A, that is, it must pass through it. Therefore, if two circles, &c Q. E. D. See N. PROP. XIII. THEOR. NE circle cannot touch another in more points than ON circle car it touches it on the infide or out fide. For, if it be poffible, let the circle EBF touch the circle ABC in more points 'than one, and firft on the infide, in the a 10.11.1. points B, D; join BD, and draw GH bifecting BD at right angles.: Therefore, because the points B, D, are in the circumfe b 2. 3. rence of each of the circles, the ftraight line BD falls within beach of them: And their centers are in the ftraight line -GI c Cor. 1. 3. which bifects BD at right angles; therefore GH puffes through the point of contacts; but it does not pafs through it, because the points B, Dare without the ftraight line GH, which is abfurd: Therefore one circle cannot touch another on the infide in more points than one. d II. 3. Nor can two circles touch one another on the outfide in more 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 the circumference of the circle ACK,, the ftraight line AC which joins them fhall fall within the circle ACK: And the circle ACK is without the circle ABC; and therefore the ftraight line AC is without this laft circle; but, because the points A, C are in the circumference of the circle ABC, the ftraight line AC must be within the fame circle, which is abfurd: Therefore one circle cannot touch another on the outfide in more than one point.: And it has been fhewn, that they cannot touch K 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 center; and thofe which are equally diftant from the center, 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 center. C Take E the center of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD: Then, because the straight line EF paffing through the center cuts the ftraight line AB, which does not pafs thro' the center, at right angles, it also bifects it: Wherefore AF is equal to FB, and AB double of AF. For the fame reason, 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 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 F b 2.30 a 3. 3. E b47. 1. EG, GC are equal to the fquare of EC: Therefore the fquares of AF, FE are equal to the fquares of CG, GE, of which Book III. which the fquare of AF is equal to the fquare of CG, because ~ AF is equal to CG; therefore the remaining square of FE is equal to the remaining fquare of EG, and the straight line FE is therefore equal to EG: But ftraight lines in a circle are faid to be equally distant from the center, when the perpendiculars 4. Def. 3. drawn to them from the center are equal: Therefore AB, CD are equally distant from the center. Next, if the straight lines AB, CD be equally diftant from the center, that is, if FE be equal to EG; AB is equal to CD: For, the fame conftruction being made, it may, as before, be demonstrated, 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. There. fore equal straight lines, &c. Q. E. D. See N. 2 20. I. HE diameter is the greateft ftraight line in a circle; Tand, of all others, that which is nearer to the cen ter is always greater than one more remote; and the greater is nearer to the center than the lefs. Let ABCD be a circle, of which the diameter is AD, and center E; and let BC be nearer to the center than FG; ÁD is greater than any straight line BC which is not a diameter, and BC greater than F From the center draw EH, EK per- er than BC. A B H E C D And, becaufe BC is nearer to the center than FG, EH EH is less than EK: But, as was demonftrated in the pre- Book III. ceeding, BC is double of BH, and FG double of FK, and then fquares of EH, HB are equal to the fquares of EK, KF, of b 5. 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 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 center 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 BII, HE are equal to the fquares of FK, KE, of which the fquare of BH is greater than the fquare of FK, becaufe BH is greater than FK; therefore the fquare of EH is lefs than the fquare of EK, and the ftraight line EH lefs than EK. Wherefore the diameter, &c. Q. E. D. Time at to the dis HE ftraight line drawn at right angles to the dia- Sec N. meter of a circle, from the extremity of it, falls without the circle; and no ftraight 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 ftraight 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 center of which is D, and the diameter AB; the ftraight 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 therefore ACD is a right angle, and he angles DAC, ACD are therefore equal to two right angles; which is impoffible: b 17, 1, F Therefore 82 be Book III. Therefore the ftraight line drawn from A at right angles to BA it may does not fall within the circle: In the fame manner, demonftrated that it does not fall upon the circumference; therefore it must fall without the circle, as AE. C 12. 1. d 19. I. F E C H A D 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 becaufe 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 lefs 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 circle, 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 small an angle it makes with AE, the circumference pafies between that ftraight line and the perAnd this is all that is to be understood, pendicular AE. 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 lefs 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 Also it is evident that there can be but one fall within it. ftraight line which touches the circle in the fame point.' € 2.3. 6 TO draw a ftraight line from a given point, either without or in the circumference, which fhall touch a given circle. First, Let A be a given point without the given circle CD; |