Book II. in ¿ 47. 1. с together with the fquare of EG, is equal to the fquare of GH But the fquares of HE, EG are equal to the fquare of GH: Therefore the rectangle BE, EF, together with the square of EG, is equal to the fquares of HE, EG: Take away the fquare of EG, which is common to both; and the remaining rectangle BE, EF is equal to the fquare of EH: But the rectangle contained by BE, EF is the parallelogram BD, becaufe EF is equal to ED; therefore BD is equal to the fquare of EH; but BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the fquare of EH: Wherefore a fquare has been made equal to the given rectilineal figure A, viz. the square defcribed upon EH. Which was to be done. THE Book III. THE ELEMENTS O F EUCL I D. E QUAL circles are thofe of which the diameters are equal, or from the centres of which the ftraight 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, fo that their centres coincide, the circles muft likewise coincide, fince the straight lines from the centres are equal. 68 Book III. VII. "The angle of a fegment is that which is contained by the "ftraight line and the circumference." VIII. An angle in a fegment is the angle con- IX. I And an angle is faid to infift or stand X. The fector of a circle is the figure contain a 10. I. b II. I. Let ABC be the given circle; it is required to find its centre. Draw within it any ftraight line AB, and bifect it in D; from the point D draw b DC at right angles to AB, and produce it to E, and bife& CE in F: The point F is the centre of the circle ABC. Fors For, if it be not, let, if poffible, G be the centre, and join Book III. GA, GD, GB: Then, becaufe DA is equal to DB, and DG C FG c 8. I. D B common to the two triangles ADG, E COR. From this it is manifeft, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bifects the other. d 10. def. I. IF any two points be taken in the circumference of a circle, the ftraight line which joins them fhall fall within the circle. Let ABC be a circle, and A, B any two points in the cir ⚫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 ci cumference, c 16. I. d 19. .1 с 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 fide is oppofited; 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 impoffible: Therefore the ftraight line drawn from A to B does not fall without the circle. In the fame manner, it may be demonstrated that it does not fall upon the circumference; it falls therefore with in it. Wherefore, if any two points, &c. Q. E. D. a 1. 3. b 8. I. IF Fa ftraight line drawn through the centre of a circle bifect a straight line in it which does not pass through the centre, it shall cut it at right angles; and, if it cuts it at right angles, it shall bifect it. Let ABC be a circle and let CD, a ftraight line drawn through the centre, bised any straight line AB, which does not pafs through the centre, in the point F: It cuts it alfo at right angles. C Take 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 fides in the one equal to two fides in the other, and the base EA is equal to the bafe EB; therefore the angle AFE is equal to the Angle BFE: But when a ftraight line ftanding upon another makes the adjacent angles equal to one another, each of them is a right 10. def, 1. angle: Therefore each of the angles AFE, BFE is a right angle; wherefore the traight line CD, drawn through the centre bifecting another AB that does not pafsthrough the centre, cuts the fame at right angles. A E F D But let CD cut AB at right angles; CD alfo bifects it, that is, AF is equal to FB. The fame conftruction 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, there |