b. 5. Def. 3. 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 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 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 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 ftraight line EH less than EK. Wherefore the diameter, &c. Q. E. D. TH PROP. XVI. THEOR, 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 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 poffible, within the circle as AC, and B draw DC to the point C where it meets the circumference. and be A D caufe DA is equal to DC, the angle DAC is equal a to the angle ACD; but DAC is a right angle, therefore ACD is a right angle, and a. 5. I. Book III. the angles DAC, ACD are therefore equal to two right angles; which is impoffible ". therefore the straight 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 cir cumference; therefore it must fall without the circle, as AE. b. 17. I. d. 19. I. And between the straight 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 F E c. 12. 1. 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 greaterdthan DG. but DA is equal to DH; therefore DH is greater than DG, thế lefs than the greater, which is impoffible. therefore no ftraight line can be drawn from the point A between AE B and the circumference, which does not cut the circle. or, which amounts to the fame thing, however great an acute angle a straight line makes with e. 2. 3. A D the diameter at the point A, or however small an angle it makes with AE, the circumference paffes between that 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 • femicircle is faid to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal angle.' 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, because if it did meet the circle in two, it would fall within Also it is evident that there can be but one ftraight line which touches the circle in the same point." it e. PROP. XVII. PROB. 10 draw a straight 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 BCD; it is required to draw a ftraight line from A which fhall touch the Book III. circle. Find the center E of the circle, and join AE; and from the a. 1. 3. center E, at the distance EA describe the circle AFG; from the b point D draw ↳ DF at right angles to EA, and join EBF, AB. b. 11. 1. AB touches 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 G E common to the two triangles AEB, FED; therefore the base DF is equal to the bafe AB, and the triangle FED to the triangle AEB, and the other angles to the other angles c. CE B A 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 center; but a straight line drawn from the extremity of C. 4. Is a diameter, at right angles to it, touches the circle . therefore AB d.Cor.16.3. 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 center E, and DF at right angles to DE; DF touches the circle d IF F a ftraight line touches a circle, the ftraight line drawn from the center to the point of contact, shall 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 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 angle; b. 17. I. and to the greater angle the greatest fide is oppofite. therefore FC c. 19. I. C a. 18. 3. See N. IF Fa ftraight line touches a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the center of the circle fhall be in that line. Let the ftraight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE; the center of the circle is in CA. For if not, let F be the center, if possible, and join CF. Because is drawn from the center to the point fore the angle FCE is equal to the B angle ACE, the lefs to the greater, that no other point which is not in A E CA, is the center; that is, the center is in CA. Therefore if a ftraight line, &c. Q. E. D. THE angle at the center of a circle is double of the THE angle at the circumference, upon the same base, that is, upon the fame part of the circumference. Let ABC be a circle, and BEC an angle at the center, and BAC Book III. an angle at the circumference, which have the fame circumference BC for their bafe; the angle BEC is double of the angle BAC. First, Let E the center of the circle be within the angle BAC, and join AE, and produce it to F. Because EA is equal to EB, the angle EAB is equal to the angle EBA; therefore the angles EAB, EBA are double of the angle EAB; but the B angle BEF is equal to the angles EAB, b EBA; therefore alfo the angle BEF is F A double of the angle EAB. for the same reason, the angle FEC is double of the angle EAC. therefore the whole angle BEC is double of the whole angle BAC. Again, Let BDC be inflected to the circumference, so that E the center of the circle be without the angle BDC, and join DE and produce it to G. It may be demonftrated, as in the first case, that the angle GECis double of the angle GDC, and that GEB a part of the firft is double of GDB a part of the other; therefore the remaining angle BEC is double of the B E remaining angle BDC. Therefore the angle at the center, &c. Q. E.D. HE angles in the fame fegment of a circle are See N. equal to one another. TH Let ABCD be a circle, and BAD, BED angles in the fame fegment BAED; the angles BAD, BED are equal to one another. Take F the center of the circle ABCD. and, first, let the segment BAED be greater than a femicircle, and join BF, FD.B and because the angle BFD is at the cen ter, and the angle BAD at the circumfe A E |