C 47. I. с Book II. together with the fquare of EG, is equal to the square of GH: But the fquares of HE, EG are equal to the fquare of GH: Therefore the rectangle BE, EF, together with the fquare of EG, is equal to the fquares of HE, EG: Take away the square 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, because 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 square has been made equal to the given rectilineal figure A, viz. the fquare defcribed upon EH. Which was to be done. ཟླ་ THE THE ELEMENTS OF EUCLI D. воок III, DEFINITIONS, I. EQUAL circles are thofe of which the diameters are equal, Book ILL or from the centres of which the ftraight lines to the cir cumferences 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 muft likewife coincide, fince the straight lines from the centres are equal.' VI. Book III. A fegment of a circle is the figure contained by a ftraight line and the cir cumference it cuts off. VII. "The angle of a fegment is that which is contained by the ftraight line and the circumference." 66 Vill. An angle in a fegment is the angle con- IX. And an angle is faid to infift or ftand upon the circumference intercepted between the straight lines that contain the angle. The fector of a circle is the figure contain ed by two ftraight lines drawn from the XI. Similar fegments of a circle, See N. a IO. I. b II. I. PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any ftraight line AB, and bifect a it in D; from the point D draw b DC at right angles to AB, and produce it to E, and bifect CE in F: The point F is the centre of the circle ABC. For, C FG c 8. 1. B D For, if it be not, let, if poffible, G be the centre, and join Book III. GA, GD, G3: Then, becaufe DA is equal to DB, and DG common to the two triangles ADG, BDG, the two fides AD, DG are equal to the two BD, DG, each to each; and the bafe GA is equal to the bafe GB, because they are drawn from the centre G *: Therefore the angle ADG is equal to the angle GDB: But when a straight line ftanding upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angled: Therefore the angle GDB is a right angle: But FDB is likewise a right angle; wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impoffible: Therefore G is not the centre of the circle ABC: In the fame manner it can be shown, that no other point but F is the centre; that is, F is the centre of the circle ABC: Which was to be found. A COR. From this it is manifeft, that if in a circle a straight line bifect another at right angles, the centre of the circle is in the line which bifects another. PROP. II. THEOR. IF any two points be taken in the circumference of a circle, the straight line which joins them fhall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference; the ftraight line drawn from A to B fhall fall within the circle. For, if it do not, let it fall, if poffible, without, as AEB; find a D the centre of the circle ABC, and join AD, DB, and produce DF, any ftraight line meeting the circumference AB to E: Then because DA is equal to DB, the angle DAB is equal b to the angle DBA; and because AE, a fide of the triangle A E 3 C D dro. Def.z: a I. 3. F E B b5.. DAE; N. B. Whenever the expreffion "ftraight lines from the centre," or "drawn " from the centre," occurs, it is to be understood that they are drawn to the circumference. C 16. I. d 19. I. с Book. III. DAE, is produced to B, the angle DEB is greater than thể 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 lefs 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. á 1. 3. b 8. I. IF PROP. III. THEOR. a ftraight line drawn through the centre of a circle bifect a straight line in it which does not pass through the centre, it fhali cut it at right angles; and, if it cuts it at right angles, it fhall bifect it. Let ABC be a circle; and let CD, a straight line drawn through the centre, bifect any straight line AB, which does not pafs through the centre, in the point F: It cuts it also 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 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 standing upon another makes the adjacent angles equal to one another, each of them is a right C10. Def..c angle: Therefore each of the angles AFE, BFE is a right angle; wherefore the ftraight line CD, drawn through the centre bifecting another AB that does not pafs through the centre, cuts the fame at right angles. d 5. I. A E F B 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 d to the angle EBF; and the right angle AFE is equal to the right angle BFE: Therefore, in the two triangles EAF, EBF, there |