than one point. for, if it be poffible, let the circle ACK touch the Book III. circle ABC in the points A, C, and join AC. therefore because the been shewn that they cannot touch on the infide in more points than one. therefore one circle, &c. Q. E. D. PKOP. XIV. THEOR. EQUAL ftraight lines in a circle are equally distant from the center; and those which are equally distant from the center, are equal to one another. Let the straight lines AB, CD in the circle ABDC be equal to one another; they are equally diftant from the center. 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 thro' the center cuts the ftraight line AB, which does not pafs thro' the center, at right angles, it also the fquare of AE is equal to the fquare angle AFE is a right angle; and for the like reafon the fquares of EG, GC are equal to the fquare of EC. therefore the fquares of AF, FE are equal to the fquares of CG, E D b. 47. 8. Book III. GE, of which the fquare of AF is equal to the fquare of CG, be, caufe AF is equal to CG; therefore the remaining fquare 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 e qually diftant from the center, when the perpendiculars drawn to them from C G F 4. Def. 3. the center are equal. therefore AB, A Sce N. . 20. X. 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 square 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. THE diameter is the greatest straight line in a circle; and of all others, that which is nearer to the center 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 From the center draw EH, EK per- F BC. a And because BC is nearer to the cen K A B H E G Ꭰ ter 73 ter than FG, EH is lefs than EK. but, as was demonftrated in Book III. the preceding, 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 which b. s. Def.8. the fquare of EH is less than the fquare of EK, because EH is lefs than EK; therefore the fquare 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 center than FG, that is, the fame construction being made, EH is less than EK. because BC is greater than FG, BH likewise 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; therefore the fquare of EH is lefs than the fquare of EK, and the straight 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 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 straight 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 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 it meets the circumference. and becaufe DA is equal to DC, the angle DAC is equal to the angle ACD; but DAC is a right angle, therefore B C D A ACD is a right angle, and the angles DAC, ACD are therefore 2. 5. x. equal Book III. equal to two right angles; which is impoffible b. therefore the ftraight line drawn from A at right angles to BA does not fall within b. 17. 1. the circle. in the fame 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 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 c. 12. 1. D draw DG perpendicular to FA, and let it meet the circumference in H. and because AGD is a right angle, and DAG lefs b than c. 2. 3. fore DH is greater than DG, the lefs the diameter at the point A, or however small an angle it makes with COR. From this it is manifeft 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. Also it is ' evident that there can be but one ftraight line which touches the ⚫ circle in the fame point.' То PROP. XVII. PROR. 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 fhall touch the Book III. circle. 2 Find the center E of the circle, and join AE; and from the cen- a. 1. 3. ter E, at the di.tance E defcribe the circle AFG; from the point D drawb DF at right angles to EA, and join EBF, AB. AB touches b. 11. 1. the circle BCD. Because E is the center 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 base DF is equal to the base AB, and the triangle FED to the triangle G CE BF 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 center; but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle. therefore AB touches the circle ; d.Cor.16.3. 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 center E, and DF at right angles to DE; DF touches the circle . PROP. XVIII. THEOR. Fa ftraight line touches a circle, the straight line drawn from the center to the point of contact, fhall be perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point C, take the center F, and draw the straight 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 angle; b. 17. 1. and to the greater angle the greateft fide is oppofite. therefore FC c. 19, 1. |