PROPOSITION XXXI.-THEOREM. In a circle, the angle in a semicircle is a right angle; but the angle in a seg ment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. LET abcd be a circle, of which the diameter is bc, and centre e; and draw ca, dividing the circle into the segments a b c a dc, and join ba, ad, dc; the angle in the semicircle bac is a right angle; and the angle in the segment a b c, which is greater than a semicircle, is less than a right angle; and the angle in the segment a dc, which is less than a semicircle, is greater than a right angle. Join a e, and produce ba to f. And because be is equal to ea, the angle ea b is equal (i. 5) to eba; also, because a e is equal to ec, the angle eac is equal to eca; wherefore the whole angle bac is equal to the two angles f abc, acb. But fac, the exterior angle of the triangle a bc, is equal (i. 32) to the two angles abc, a cb; therefore the angle bac is equal to the angle fac, and each of them is d therefore a right angle (i. def. 10). Wherefore the angle bac in a semicircle is a right angle. And because the two angles a b c, bac of bk the triangle a b c are together less (i. 17) than two right angles, and that bac is a right angle, a b c must be less than a right angle ; and therefore the angle in a segment a b c greater than a semicircle, is less than a right angle. And because a b c d is a quadrilateral figure in a circle, any two of its opposite angles are equal (iii. 22) to two right angles: therefore the angles abc, adc are equal to two right angles; and a b c is less than a right angle; wherefore the other adc is greater than a right angle. Besides, it is manifest that the circumference of the greater segment abc falls without the right angle cab; but the circumference of the less segment adc falls within the right angle caf. And this is all that is meant, when in the Greek text, and the translations from it, the angle of the greater segment is said to be greater, and the angle of the less segment is said to be less, than a right angle. CoR. From this it is manifest that if one angle of a triangle be equal to the other two it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal they are right angles. e a PROPOSITION XXXII.—THEOREM. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. LET the straight line ef touch the circle a bed in b, and from the point b let the straight line bd be drawn, cutting the circle. The angles which bd makes with the touching line ef shall be equal to the angles in the alternate segments of the circle: that is, the angle fbd is equal to the angle which is in the segment dab, and the angle dbe to the angle in the segment b c d. From the point b draw (i. 11) ba at right angles to ef, and take any point c in the circumference bd, and join a d, dc, cb; and because the straight line ef touches the circle abcd in the point b, and ba is drawn at right angles to the touching line from the point of contact b, the centre of the circle is in ba (iii. 19); therefore the angle adb in a semicircle is a right (iii. 31) angle, and consequently the other two angles bad, abd are equal (i. 32) to a right angle. But a bf is likewise a right angle: therefore the angle a bf is equal to the angles bad, a bd. Take from these equals b f the common angle a bd; therefore the remain ing angle dbf is equal to the angle bad, which is in the alternate segment of the circle ; and because abcd is a quadrilateral figure in a circle, the opposite angles bad, bcd are equal to two right angles (iii. 23): therefore the angles dbf, d be, being likewise equal (i. 13) to two right angles, are equal to the angles bad, bod: and dbf has been proved equal to bad. Therefore the remaining angle dbe is equal to the angle bed in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D. PROPOSITION XXXIII.- PROBLEM. Upon a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle. LET a b be the given straight line, and the angle at c the given rectilineal angle; it is required to describe upon the given straight line a b a segment of a circle, containing an angle equal C to the angle c. First, let the angle at c be a right angle, and bisect (i. 10) a b in f, and from the centre f, at the distance fb, describe the semicircle a hb; f b therefore the angle a hb in a semicircle is (iii. 31) equal to the right angle at c. But if the angle c be not a right angle, at the point a, in the straight line ab, make (i. 23) the angle bad equal to the angle c, and from the point a draw (i. 11) a e at right angles to ad; bisect (i. 10) a b in f, and from f draw (i. 11) fg at right angles to a b, and join gb: and because a f is equal to fb, and fg common to the triangles afg; bfg, the two sides a f, fg are f equal to the two bf, fg; and the angle a fg is equal to the angle bfg; therefore the base ag is equal (i. 4) to the base gb: and the circle described from the centre g, at the distance ga, shall pass through the point b; let this be the circle a hb: and because from the point a, the extremity of the diameter a e, h a d is drawn at right angles to a e, therefore ad (iii. 16. cor.) touches the circle ; с f and because a b drawn from the point of a contact a cuts the circle, the angle dab is equal to the angle in the alternate segment a hb (iii. 32): but the angle d a b is equal to the angle c, therefore also the angle c is d equal to the angle in the segment abb: wherefore upon the given straight line a b the segment a hb of a circle is described, which contains an angle equal to the given angle at c. Which was to be done. g PROPOSITION XXXIV.—PROBLEM. To cut off a segment from a given circle, which shall contain an angle equal to a given rectilineal angle. LET a b c be the given circle, and d the given rectilineal angle; it is required to cut off a segment from the circle a b c that shall contain an angle equal to the given angle d. Draw (iii. 17) the straight line e f touching the circle a b c in the point b, and at the point b, in the straight line bf, make (i. 23) the angle fb c equal to the angle d: therefore, because the straight line ef touches the circle a b c, and b c is drawn from the point of contact b, the angle fb c is equal (iii. 32) to the angle in the alternate segment d bac of the circle : but the angle fb c is equal to the angle d : therefore the angle f in the segment bac is equal to the angle b d: wherefore the segment bac is cut off from the given circle a b c, containing an angle equal to the given angle d. Which was to be done, L PROPOSITION XXXV.-THEOREM. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. LET the two straight lines a c, b d, within the circle a b c d, cut one an other in the point e; the rectangle contained by a e, ec is equal to the rectangle contained by be, ed. d If ac, bd pass each of them through the centre, so that e is the centre, it is evident that a e, ec, be, ed being all equal, the rectangle a e, e c, is likewise equal to the rectangle be, ed. b But let one of them bd pass through the centre, and cut the other a c which does not pass through the centre, at right angles, in the point e : then, if bd be bisected in f, f is the centre of the circle abcd; join a f. And because bd, which passes through the centre, cuts the straight line a c, which does d not pass through the centre, at right angles in e; a e, ec are equal (iii. 3) to one another : and because the straight line bd is cut into two equal parts in the point f, and into two unequal in the point e, the rectangle be, ed, together, with the square of ef, is equal (ii. 5) to the square of fb; that is, to the square of fa: but, the squares of a e, ef, are equal (i. 47) to the square of fa : therefore the rectangle be, ed, b together with the square of ef, is equal to the a e, ef: take away the common square of ef, and the remaining rectangle be, ed is equal to the remaining square of a e; that is, to the rectangle ae, ec. Next, let bd, which passes through the centre, cut the other a C, which does not pass through the centre, in e, but not at right angles : then, as before, if bd be bisected in f, f is the centre of the circle. Join af, and from f draw (i. 12) fg perpendicular to ac; therefore ag is equal (iii. 3) to ge; wherefore the rectangle a e, ec, together with the square eg is equal (ii. 5) to the square of a g: to each of these equals add the square of gf; therefore the rectangle a e, ec, together with the squares of eg, gf, is equal to the squares of a 8; gf: but the squares of eg, gf, are equal (i. 47) to the square of ef; and the squares of ag, & f, are equal to the square of af: therefore the rectangle a e, ec, c together with the square of ef, is equal to the square of a f; that is, to the square of fb: but b the square of fb is equal (ii. 5) to the rectangle be, e d, together with the square of ef; therefore the rectangle a e, ec, together with the square of ef, is equal to the rectangle be, ed, together with the square of e f: take away the common square of ef, and the remaining rectangle a e, ec, is therefore equal to the remaining rectangle be, ed. squares of 09 Lastly, let neither of the straight lines ac, bd pass through the centre : take the centre f, and through e, the intersection of the straight lines ac, db, draw the diameter gefh: and because the rectangle a e, ec, is equal, as has been shewn, to the rectangle ge, eh; and, for the same reason, the rectangle be, ed is equal to the same rectangle ge, eh; therefore the rectangle a e, ec is equal to the rectangle be, ed. Wherefore, if two straight lines, &c. Q. E. D. PROPOSITION XXXVI.—THEOREM. If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it, the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it. LET d be any point without the circle a b c, and d cald b two straight lines drawn from it, of which dca cuts the circle, and db touches the same : the rectangle a d, dc is equal to the square of d b. Either dca passes through the centre, or it does not; first, let it pass through the centre e, and join eb; therefore the d angle ebd is a right (iii. 18) angle; and because the straight line a c is bisected in e, and produced to the point d, the rectangle ad, dc, together with the square of ec, is equal (i. 6) to the square of ed, and ce is equal to eb: therefore the rectangle ad, dc, together with the square of eb, is equal to b the square of ed. But the square of ed is equal (i. 47) to the squares of e b, bd, because ebd is a right angle: therefore the rectangle a d, dc, together with the square of e b, is equal to the squares of e b, bd. Take away the common square of eb; therefore the remaining rectangle a d. dc, is equal to the square of the tangent d b. a But if d ca does not pass through the centre of the circle a bc, take (iii. 1) the centre e, and draw ef perpendicular d (i. 12) to a c, and join eb, ec, ed : and because the straight line e f, which passes through the centre, cuts the straight line a C, which does not pass through the centre, at right angles, it shall likewise bisect (iii. 3) it; therefore a f is equal b. to fc: and because the straight line ac is bisected in f, and produced to d, the rectangle a d, dc, together with the square offc, is equal (ii. 6) e to the square offd. To each of these equals add the square of fe; therefore the rectangle a d, dc, together with the squares of cf, fe, is equal to the squares of d f, fe; but the square of e d is equal |