† 15 Def. 1. 20. 1. #5 Def. 3. *15 Def. 1. + 5. 1. + Hyp. 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 also AD is greater than BC. And, because BC is nearer to the centre than FG, EH is less than EK: But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares EK, KF, of which the 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 KF; 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. PROPOSITION XVI. THEOR.-The straight line drawn at right angles to the diameter 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, 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 C, where it meets the circumference: And because DA is equal* to DC, the angle DAC is equal to the angle ACD: but DAC is a right‡ angle; therefore ACD is a right angle, and the angles DAC, 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. B Ꭰ 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 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 less* than a right angle, DA is greater† than DG: B FE H A But DA is equal‡ to DH; therefore DH is greater than DG, the less than the greater, which is impossible: therefore no straight line can be drawn from the point A between AE and the circumference, which does not cut the circle; or, which amounts to the same 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 AE. '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.' Q. E. D. || 17. 1. 12. 1. Constr. * 17. 1. † 19. 1. 15 Def. !. 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 fall within it. Also, it is evident || 2. 3. that there can be but one straight line which touches the circle in the same point.' PROPOSITION XVII. PROB. To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. * 1. 3. † 11. 1. First, let A be a given point without the given circle BCD; it is required to draw a straight line from A which shall touch the circle. Find* the centre E of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw† DF at right angles to EA, and join EBF, AB: AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, ‡ 15 Def. 1. EA is equal to EF, and ED to # 4. 1. EB; Therefore the two sides 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 CE B to the triangle AEB, and the other angles to the other angles; 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 centre: but a straight line drawn from the extremity of a diameter, at right § Cor. 16. 3. angles to it, touches the circle§: therefore AB touches the circle; and it is drawn from the given point A. was to be done. Which But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF ¶ Cor. 16. 3. at right angles to DE: DF touches the circle ¶. * 17. 1. † 19. 1. PROPOSITION XVIII. THEOR.-If a straight line touch a circle, the straight line drawn from the centre 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 centre F, and draw the straight line FC: FC is per- For if it be not, from the point And because FGC is a right angle, GE But FC is equal to FB; therefore FB is greater than 15 Def. 1. FG, the less than the greater, which is impossible: wherefore FG is not perpendicular to DE. In the same manner it may be shown, that no other is perpendicular to it besides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D. PROPOSITION XIX. THEOR.-If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line. Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE; the centre of the circle is in CA. For, if not, let F be the centre, if possible, and join CF. ABC, and FC is drawn from the But ACE is also a right angle; therefore the angle FCE is equal B to the angle ACE, the less to the greater, which is impossible; wherefore F is not the centre of the circle ABC. D In the same manner it may be shown, that no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore, if a straight line, &c. PROPOSITION XX. Q. E. D. THEOR.-The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference. Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base; the angle BEC is double of the angle BAC. First, let E, the centre of the circle, be within the angle BAC, and join AE, and produce it to F: 18. 3. * 5.1. † 32. 1. 1. 3. + 20. 3. Because EA is equal to EB, the angle EAB is equal* But the angle BEF is equal to the 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. B E Again, let E, the centre of the circle, be without the angle BDC, and join DE, and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double of the angle GDC, and that GEB, a part of the first, is double of GDB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BDC. Therefore the angle at the centre, &c. Q. E. D. PROPOSITION XXI. E THEOR.-The angles in the same segment of a circle are equal to one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED are equal to one another. Take* F, the centre of the circle ABCD: and, first, let the segment BAED be greater than a semicircle, and join BF, FD: And because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. BCD for their base; therefore the angle BFD is doublef of the angle B BAD: For the same reason, the angle BFD is double of the angle BED; F Therefore the angle BAD is equal to the angle BED. |