.. 4 ABC < a right angle. Hence Prop. 32. And therefore each of them is a right angle (I. Def. 10); Therefore the angle in a semicircle BAC is a right angle. And because the two angles ABC, BAC, of the triangle ABC, are together less than two right angles (I. 17), and that BAC has been shown to be a right angle; Therefore the angle ABC is less than a right angle. Therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. And, because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles (III. 22); Therefore the angles ABC, ADC are together equal to two right angles. But the angle ABC has been shown to be less than a right angle; Therefore the angle ADC is greater than a right angle; Therefore the angle in a segment ADC, less than a semicircle, is greater than a right angle. Therefore, the angle, &c. Q.E.D. COROLLARY.-From this demonstration it is manifest that, if one angle of a triangle be equal to the other two, it is a right angle. For the angle adjacent to it is equal to the same two angles (I. 32). And, when the adjacent angles are equal, they are right angles (I. def. 10). Proposition 32.-Theorem. The angles contained by a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segments of the circle. Let EF be a tangent to the circle ABCD, and BD a chord drawn from the point of contact B, cutting the circle. The angles which BD makes with the tangent EF shall be equal to the angles in the alternate segments of the circle; That is, the angle DBF shall be equal to the angle in the segment BAD, and the angle DBE shall be equal to the angle in the segment BCD. CONSTRUCTION. From the point B draw BA at right angles to EF (I. 11). Take any point C in the circumference BD, and join AD, DC, CB. PROOF. Because the straight line EF touches the circle ABCD at the point B (Hyp.), and BA is drawn at right angles to the tangent from the point of contact B (Const.), The centre of the circle is in BA (III. E 19). Therefore the angle ADB, being in a semicircle, is a right. ADB is angle (III. 31). Therefore the other two angles BAD, ABD are equal to right angle (I. 32). But ABF is also a right angle (Const.); a right angle, a and 4 BAD + < ABD = a right angle = ABF. Therefore the angle ABF is equal to the angles BAD, ABD. From each of these equals take away the common angle ABD ; Therefore the remaining angle DBF is equal to the re-.. BAD maining angle BAD, which is in the alternate segment of the circle (Ax. 3). = 4 DBF, And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are together equal to two Also, right angles (III. 22). 4 BCD + < BAD= But the angles DBF, DBE are together equal to two 2 right right angles (I. 13); Therefore the angles DBF, DBE are together equal to the angles BAD, BCD. And the angle DBF has been shown equal to the angle BAD; Therefore the remaining angle DBE is equal to the angle BCD, which is in the alternate segment of the circle (Ax. 3). Therefore, the angles, &c. Q.E.D. Proposition 33.-Problem. Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and C the given rectilineal angle, angles = DBF + DBE. .. 4 DBE = 4 BCD, Angle in a semicircle It is required to describe, on the given straight line AB, a segment of a circle, containing an angle equal to the angleC. CASE I. Let the angle C be a right angle, CONSTRUCTION,-Bisect AB in F (I. 10), From the centre F, at the distance FB, describe the semicircle AHB. Then AHB shall be the segment required. PROOF.-Because AHB is a semicircle, the angle AHB is a right in it is a right angle, and therefore equal to the angle C (III. 31). angle. At point A make BAD = C; CASE II.-Let C be not a right angle. CONSTRUCTION.-At the point A, in the straight line AB, make the angle BAD equal to the angle C (I. 23). and draw AE at right angles to AD. From F, From the point A draw AE at right angles to AD (I. 11). From the point F draw FG at right angles to AB (I. 11), middle of and join GB. AB, draw perpendicular, meeting AE in G. Because AF is equal to BF (Const.), and FG is common to the two triangles AFG, BFG; The two sides AF, FG are equal to the two sides BF, FG, each to each; And the angle AFG is equal to the angle BFG (Const.); Therefore the base AG is equal to the base BG (I. 4). And the circle described from the centre G, at the distance GA, will therefore pass through the point B. Let this circle be described; and let it be AHB. The segment AHR shall contain an angle equal to the given rectilineal angle C. PROOF.-Because from the point A, the extremity of the liameter AE, AD is drawn at right angles to AE (Const.); Therefore AD touches the circle (III. 16, cor.) Because AB is drawn from the point of contact A, the angle DAB is equal to the angle in the alternate segment AйB (III. 32). And AD the circle, touches and .. 4 But the angle DAB is equal to the angle C (Const.); Therefore the angle in the segment AHB is equal to the in AHB = angle C (Ax. 1). Therefore, on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C. Q.E.F. Proposition 34.-Problem. From a given circle to cut off a segment which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle. It is required to cut off from the circle ABC a segment that shall contain an angle equal to the angle D. CONSTRUCTION. < DAB or C. gent EBF, Draw the straight line EF touching the Draw tancircle ABC in the point B (III. 17); And at the point B, in the straight line BF, make the angle FBC equal to the angle D (I. 23). Then the segment BAC shall contain an angle equal to the given angle D. B PROOF.-Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B (Const.); Therefore the angle FBC is equal to the angle in the alternate segment BAC of the circle (III, 32). But the angle FBC is equal to the angle D (Const.); and given 2. Therefore the angle in the segment BAC is equal to the BAC angle D (Ax, 1). Therefore, from the given circle ABC, the segment BAC has been cut off, containing an angle equal to the given angle D. Q.E.F. = 4 FBC = 4 D. Proposition 35.-Theorem. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them shall be equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD cut one another in the point E, within the circle ABCD. B The rectangle contained by AE and EC shall be equal to the rectangle contained by BE and ED. CASE I. Let AC, BD pass each of them through the centre. PROOF.-Because E is the centre, EA, EB, EC, ED are all equal (I. def. 15); Therefore the rectangle AE, EC is equal to the rectangle BE, ED. CASE II.-Let one of them, BD, pass through the centre, and cut the other, AC, which does not pass through the centre, at right angles, in the point E. AE EC. BE⚫ED + FB2 AF2 = ...BE ED CONSTRUCTION.-Bisect BD in F, then F is the centre of the circle; join AF. PROOF.-Because BD, which passes through the centre, cuts AC, which does not pass through the centre, at right angles in E (Hyp.); Therefore AE is equal to EC (III. 3). And because BD is cut into two equal parts in the point F, and into two unequal parts in the point E, The rectangle BE, ED, together with the square on EF, is equal to the square on FB (II. 5); that is, the square on AF. But the square on AF is equal to the squares on AE, EF (I. 47); } Therefore the rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF (Ax. 1). Take away the common square on EF; Then the remaining rectangle BE, ED is equal to the =AE= remaining square on AE; that is, to the rectangle AE, EC, since AE is equal to EC. AE EC. |