a Take e the centre of the circle a bdc, and from it draw ef, eg, perpendiculars to a b, cd. Then, because the straight line e f, passing through the centre, cuts the straight line a b, which does not pass through the centre, at right angles, it also bisects it (iii. 3); wherefore a f is equal to fb, and a b double of af. For the same reason, cd is double of cg; and a b is equal to cd; therefore 8 afis equal to cg: And because a e is equal to ec, f the square of a e is equal to the square of ec; but the squares of a f, fe, are equal (i. 47) to the square d of a e, because the angle a fe is a right angle; and, for the like reason, the squares of eg, g c, are equal b to the square of ec; therefore the squares of a f, 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 a b c d must be equally distant from the centre. Next, if the straight lines a b c d be equally distant from the centre, that is, if fe be equal to eg; a b is equal to cd. For, the same construction being made, it may, as before, be demonstrated, that a b is double of a f, 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 a f is equal to the remaining square of cg; and the straight line a f is therefore equal to cg: and a b is double of a f, 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 a diameter is a d, and the centre e; and let bc be nearer to the centre than fg; ad is greater than any straight line b c, which f is not a diameter, and b c greater than fg. From the centre e, draw eh, ek, perpendiculars to bc,fg, and join e b, ec, ef; h and because a e is equal to e b, and ed to ec, a d is equal to eb, ec; but eb, ec are greater than b c (i. 20); wherefore, also, ad is greater than b c. And, because b c is nearer to the centre 8 than fg, eh is less (iii. def. 5) than ek. But, as was demonstrated in the preceding, d b.c is double of bh, and fg double of fk, 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 e h 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 be 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, e h 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. extremity of it, falls without the circle ; and no straight line can be angles to it, as not to cut the circle. LET a b c be a circle, the centre of which is d, and the diameter a b: the straight line drawn at right angles to a b from its extremity a, shall fall without the circle. For, if it does not, let it fall, if possible, within the circle, as a c, and draw dc to the point c where it meets the circumference. And because da is equal to dc, the angle dac is b equal (i. 5) to the angle acd; but dac is a right angle, therefore a cd is a right angle, and 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 a e. And between the straight line a e and the f, e 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 to fa, and let it meet the circumference in h; b and because a gd is a right angle, and dag less (i. 17) than a right angle, da is greater (i. 19) 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 a e, 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 e bf, ab: a b touches the circle bed. Because e is the centre of the circles bcd, afg; ea is equal to ef, and ed to eb; therefore the two sides a e, e b are equal to the two fe, ed, and they contain the angle at e common & lc to the two triangles a e b, fed: therefore the base d f is equal to the base a b; and the triangle fed to the triangle a eb, and the other angles to the other angles (i. 4): therefore the angle e ba is equal to the angle edf; but edf is a right angle, wherefore e ba is a right angle ; and e b 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 a b 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 a b c in the point c; take the centre f, and draw the straight line fc; fc is perpendicular to de. a For, if it be not, from the point f draw fbg perpendicular to de; and because fgc is a right angle, g cf(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 per(b pendicular to de. In the same manner it may be shewn, that no other is perpendicular to it d с ge besides fc, that is, fc is perpendicular to de. Therefore, if a straight line, &c. Q. E. D. PROPOSITION XIX.—THEOREM. a 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 a b c 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; because de touches the circle a b c, 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. But ace is also bl a right angle; therefore the angle fce is equal to the angle a ce, the less to the greater, which is impossible : wherefore f is not the centre of the circle a b c. In the same manner, it may be shewn, that no other point which is not in c a, d is the centre ; that is, the centre is in ca. Therefore, if a straight line, &c. Q. E. D. PROPOSITION XX.—THEOREM. B 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 a b c 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 ba c, and join a e, and produce it to f: because ea is equal to eb, the angle e a b is equal (i. 5) to the angle eba; therefore the angles e a b, e ba are double of the angle e ab; but the angle bef is equal (i. 32) to the angles e a b, e ba; therefore с also the angle bef is double of the angle ea b. f 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 e, the centre of the circle, be without d the angle bd c, and join d e, 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 bd c. Therefore the angle at the centre, &c. Q. E. D. a PROPOSITION XXI.—THEOREM. The angles in the same segment of a circle are equal to one another. LET a b c d be a circle, and bad, bed angles in the same segment ba ed: the angles bad, bed are equal to one another. Take f the centre of the circle abcd: and, first, let the segment ba ed be greater than a semicircle, and join b f, 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. bod for their base ; by d therefore the angle bfd is double (iii. 20) of the angle ba d. For the same reason, the angle bfd is double of the angle bed: therefore the angle bad is equal to the angle bed. |