Book III. PROP. XXXI. THEOR. In a circle, the angle in a semicircle is a right angle; but the angle in a segment 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 ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. a A Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA; also because AE a 5. 1. is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB: But FAC, the exterior angle of the triangle ABC, is also equal to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right angle: Wherefore the angle BAC in a semicircle is a right angle. d B F D b 32. 1. C E c 10. Def. 1. And because the two angles ABC, BAC of the triangle ABC are together less than two right angles, and BAC is d 17. 1. a right angle, ABC must be less than a right angle; and 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 equal to two right angles; e 22. 3. therefore the angles ABC, ADC are equal to two right angles; and ABC 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 Book III. 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.' a 11. 1. b 19. 3. 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. PROP. XXXII. THEOR. If a straight line touches a circle, and from the point of contact a straight line is drawn cutting the circle, the angles made by this line with the line touching the circle, are equal to the angles in the alternate segments of the circle. Let the straight line EF touch the circle ABCD 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 are 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 BCD. A From the point B draw a BA at right angles to EF, and take any point C in the circumference BD, and join AD, 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; therefore the angle ADB in a semicircle is a right angle, and consequently the other two and 32. 1. gles BAD, ABD are equal d to a right angle: But ABF is likewise a right angle; therefore the angle ABF is equal to the c 31. 3. e 22. 3. angles BAD, ABD: Take from E B ABD, and there will remain the F angle DBF 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 equale to two right angles; therefore the angles DBF, DBE, being f likewise equal to two right angles, are equal to the angles Book III. BAD, BCD; and DBF has been proved equal to BAD: ~ Therefore the remaining angle DBE is equal to the angle f 13. I. BCD in the alternate segment of the circle. Wherefore, "if a straight line," &c. Q. E. D. PROP. XXXIII. PROB. Upon a given straight line to describe a segment of See N. a circle containing an angle equal to a given rectili neal angle. Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle containing an angle equal to the angle C. First, Let the angle at C be a right angle, bisect AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; the angle AHB being in a semicircle, is equal to the right angle at C. H a 10. 1. A F B b 31. 3. But if the angle C be not a right angle, at the point A, in the straight line AB, make the angle BAD equal to the angle c 23. 1. C C, and from the point A draw AE at right angles to AD; bisect AB in F, and from F draw FG at right angles to AB, and join GB: And because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG; and the angle AFG is also equal to the angle BFG; therefore the base AG is equal to the base GB: and the circle described from the centre G, at the distance GA, will pass through the point B; let this be the circle AHB: And because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, there e 4. 1. AHB: Wherefore, upon the given straight line AB, a segment AHB of a circle is described, which contains an angle equal to the given angle at C. Which was to be done. a 17. 3. b 23. 1. c 32. 3. PROP. XXXIV. PROB. -To cut off a segment from a given circle 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 a segment from the circle ABC that shall contain an angle equal to the given angle D. A Draw a the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF, make the angle FBC equal to the angle D: Therefore, because the straight line EF touches the circle ABC, D and BC is drawn from to the angle in the alter nate segment BAC: But C the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D: Wherefore the segment BAC is cut off from the given circle ABC, containing an angle equal to the given angle D. Which was to be done. Book III. PROP. XXXV. THEOR. If two straight lines within a circle cut one another, See N. 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 AC, BD within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC is equal to the rectangle con tained by BE, ED. If AC, BD pass each of them through the centre, so that E is the centre; it is evident, that AE, EC, BE, ED, being all equal, the rectangle AE, EC is equal B to the rectangle BE, ED. Ꭰ But 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: Then, if BD be bisected in F, F is the centre of the circle ABCD; join AF: and because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AE, EC are equal a 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 A equal to the square of FB; that is, to the square of FA; but the squares of AE, EF are equal to the square of FA; therefore the rectangle BE, a 3. 3. F E b 5. 2. B ED, together with the square of EF, is equal to the squares of AE, EF: Take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC. Next, Let BD, which passes through the centre, cut the other AC, 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 FG per d c 47. 1. d 12. 1. |