Take e the centre of the circle abdc, and from it draw ef, eg, perpendiculars to ab, cd. Then, because the straight line e f, passing through the centre, cuts the straight line ab, which does not pass through the centre, at right angles, it also bisects it (iii. 3); wherefore a f is equal to fb, and ab double of af. For the same reason, ed is double of cg; and ab is equal to cd; therefore afis equal to cg. And because ae is equal to ec, the square of ae is equal to the square of ec; but the squares of af, fe, are equal (i. 47) to the square of a e, because the angle afe is a right angle; and, for the like reason, the squares of eg, gc, are equal to the square of ec; therefore the squares of af, a e d b fe, are equal to the squares of cg, ge, of which the square of af is equal to the square of cg, because af is equal to cg; therefore the remaining square of fe is equal to the remaining square of eg, and the straight line ef is therefore equal to eg. But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal (iii. def. 4). Therefore ab, cd must be equally distant from the centre. Next, if the straight lines ab, cd be equally distant from the centre, that is, if fe be equal to eg; abis equal to cd. For, the same construction being made, it may, as before, be demonstrated, that ab is double of af, and cd double of cg, and that the squares of ef, fa are equal to the squares of eg, gc; of which the square of fe is equal to the square of eg, because fe is equal to eg; therefore the remaining square of af is equal to the remaining square of cg; and the straight line a f is therefore equal to cg: and ab is double of af, and cd double of cg; wherefore ab is equal to cd. Therefore equal straight lines, &c. Q. E. D. PROPOSITION XV. - THEOREM. The diameter is the greatest straight 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 be a circle, of which the diameter is ad, and the centree; and let bc be nearer to the centre than fg; ad is greater than any straight line bc, which is not a diameter, and be greater than fg. From the centre e, draw eh, ek, perpendiculars to b c, fg, and join e b, e c, ef; and because ae is equal to eb, and ed to ec, ad is equal to eb, ec; but eb, ec are greater than bc (i. 20); wherefore, also, ad is greater than bc. And, because bc is nearer to the centre than fg, eh is less (iii. def. 5) than ek. But, as was demonstrated in the preceding, bc is double of bh, and fg double of fk, f ab h e C d and the squares of eh, hb are equal to the squares of 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 be 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 fk; 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. - THEOREM. 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 b d C 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 a equal (i. 5) to the angle acd; but dac is a right angle, therefore acd is a right angle, and the angles dac, acd are therefore equal to two right angles, which is impossible (i. 17). Therefore the straight line drawn from a at right angles to ba, does not fall within the circle. 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. e 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 (i. 12) dg perpendicular to fa, and let it meet the circumference in h; and because agd is a right angle, and dag less a (i. 17) than a right angle, da is greater (i. 19) than dg. 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 a e. 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. 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 (iii. 2). Also, it is evident that there can be but one straight line which touches the circle in the same point. PROPOSITION XVII.—PROBLEM. To 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 shall touch the circle. Find (iii. 1) 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 (i. 11) df at right angles to ea, and join ebf,ab:ab touches the circle bcd. d a Because e is the centre of the circles bcd, afg; ea is equal to ef, and ed to 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 dfis equal to the base ab; and the triangle fed to the triangle aeb, and the other angles to the other angles (i. 4): therefore the angle eba is equal to the angle edf; butedf 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 angles to it, touches the circle (iii. 16. cor.): therefore 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 dfat right angles to de; df touches the circle (iii. 16. cor.). PROPOSITION XVIII.—THEOREM. If a straight line touches 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; fcis 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 (i. 17) is an acute angle; and to the greater angle the greatest (i. 19) side is opposite: therefore fc is greater than fg; but fc is equal to fb; therefore fb is greater than fg, the less than the greater, which is impossible: wherefore fg is not perpendicular to de. In the same manner it may be shewn, that no other is perpendicular to it besides fc, that is, fc is perpendicular to de. b ge Therefore, if a straight line, &c. Q. E. D. PROPOSITION XIX. - THEOREM. If a straight line touches 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 inc; and from clet 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; because de touches the circle abc, and fc is drawn from the centre to the point of contact, fc is perpendicular (iii. 18) to de; therefore fce is a right angle. Butace is also a right angle; therefore the angle fce is equal to the angle ace, the less to the greater, which is impossible: wherefore f is not the centre of the circle abc. In the same manner, it may be shewn, 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. Q. E. D. PROPOSITION XX. - THEOREM. 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 becan angle at the centre, and bacan angle at the circumference, which have the same circumference bc for their base; the angle bec is double of the angle bac a First, let e, the centre of the circle, be within the angle bac, and join a e, and produce it to f: because ea is equal to eb, the angle eab is equal (i. 5) to the angle eba; therefore the angles eab, eba are double of the angle eab; but the angle bef is equal (i. 32) to the angles eab, eba; therefore also the angle bef is 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 e b f angle bac. a 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 d e of the other; therefore the remaining angle bec is double of the remaining angle.bdc. Therefore the g C angle at the centre, &c. Q. E. D. b PROPOSITION XXI. - THEOREM. 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. bed for their base; b therefore the angle bfd is double (iii. 20) of the angle bad. For the same reason, the angle bfd is double of the angle bed: therefore the angle bad is equal to the angle bed. a e C f d |