PROP. XIV. PROB. To describe a square that shall be equal to a given rectilineal figure. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. H Describe a the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is now done: But if they are not A b c EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, describe the semicircle BHF; and produce DE to H, and join GH: Therefore because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal to the square of GF: But GF is equal to GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: But the squares of HE, EG are equal to the square of GH: Therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG: take away the square of EG, which is common to both; and the remaining rectangle BE, EF is equal to the square of EH: but the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH: but BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. be done. Which was to THE ELEMENTS OF EUCLID. BOOK III. DEFINITIONS. I. EQUAL circles are those of which the diameters are equal, or Book III. from the centres of which the straight lines to the circumferences are equal. This is not a definition but a theorem, the truth of which is evident; for if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, 'since the straight lines from their centres are equal.' "The angle of a segment is that which is contained by the "straight line and the circumference." VIII. An angle in a segment is the angle con- IX. And an angle is said to insist or stand X. The sector of a circle is the figure contain- See N. a 10. 1. b 11. 1. XI. Similar segments of a circle. PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. b Draw within it any straight line AB, and bisect it in D; from the point D draw DC at right angles to AB, and produce it to E, and bisect CE in F: The point F is the centre of the circle ABC. For, if it be not, let, if possible, G be the centre, and join Book III. GA, GD, GB: Then because DA is equal to DB, and DG F C G c 8. 1. B ID E common to the two triangles ADG, d COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. PROP. II. THEOR. If any two points be taken in the circumference of a circle, the straight line which joins them must fall within the circle. Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B must fall within the circle. For, if it do not, let it fall, if possible, without, as AEB; finda D the centre of the circle ABC, and join AD, DB, and produce DF, any straight line meeting the circumference AB, to E: Then because DA is equal to DB, the angle DAB is equal to the angle DBA; and because AE, a side of the triangle d 10. Def. 1. C D a 1. 3. F A A E E B b 5. 1. N. B. Whenever the expression "straight lines from the centre," or "drawn from the centre," occurs, it is to be understood that they are drawn to the circumference. c 16. 1. d 19. l. Book III. DAE, is produced to B, the angle DEB is greater than the angle DAE; but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: but to the greater angle the greater side is opposite; DB is therefore greater than DE: But DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impossible: Therefore the straight line drawn from A to B does not fall without the circle. In the same manner, it may be demonstrated, that it does not fall upon the circumference: it falls therefore within it. Wherefore, "if any two points," &c. Q. E. D. a 1. 3. PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and if it cut it at right angles it will bisect it. Let ABC be a circle; and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It also cuts it at right angles. Take a E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. But when a straight line standing upon another, makes the adjacent angles equal to one another, each of them is c 10. Def. 1. a right angle: Therefore each of the b 8. 1. d 5. 1. angles AFE, BFE is a right angle; E But let CD cut AB at right angles; The same construction being made, because EA, EB from the centre are equal to one another, the angle EAF is equal to the angle EBF; and the right angle AFE is equal to the right angle BFE: Therefore, in the two triangles, EAF, EBF, d |