Book II. of HE, EG are equal C. 47. I. H to the fquare of GH. therefore the rec→ tangle BE, EF together with the fquare 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 e qual to the square of EH. but the rectan A G E gle 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 fquare has been made equal to the given rectilineal figure A, viz. the fquare defcribed upon EH. Which was to be done. Book III. E THE ELEMENTS OF EUCL I D. BOOK III. DEFINITIONS. I. QUAL circles are thofe of which the diameters are equal, or from the centers 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, so that ⚫ their centers coincide, the circles must likewise coincide, fince the ftraight lines from their centers are equal.' II. A ftraight line is faid to touch a circle, when it meets the circle and being produced does not cut it. III. Circles are faid to touch one another, which meet but do rot cut one another. IV. Straight lines are faid to be equally diftant from the center of a circle, when the perpendiculars drawn to them from the center are equal. V. And the ftraight line on which the greater perpendicular falls, is faid to be farther from the center. Book III. 2. IO. I. b. II. 1. VI. A fegment of a circle is the figure con- VII. "The angle of a segment is that which is contained by the ftraight "line and the circumference." VIII. An angle in a segment is the angle con- IX. And an angle is faid to infift or stand X. The fector of a circle is the figure con- XI. Similar fegments of a circle, PROP. I. PROB. To find the center of a given circle. Let ABC be the given circle; it is required to find its center. Draw within it any straight line AB, and bisect a it in D; from the point D drawb DC at right angles to AB, and produce it to E, and bifect CE in F. the point F is the center of the circle ABC. For if it be not, let, if poffible, G be the center, and join GA, Book III. GD, GB. then because 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 base GB, because they are drawn from the center G*. therefore the angle ADG is equal F .c. . I. B ID E to the angle GDB. but when a ftraight line standing upon another straight line, makes the adjacent angles equal to one another, each of the angles is a right angle d. 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 lefs, which is impoffible. therefore G is not the center of the circle ABC. in the fame manner it can be shewn, that no other point but F is the center; that is, F is the center of the circle ABC. Which was to be found. COR. From this it is manifeft, that if in a circle a straight line bisect another at right angles, the center of the circle is in the line which bifects the other. d.io.Def.1. [F 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 circumference; the straight 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 center of the circle ABC, and join AD, DB, and produce DF any straight line meeting the circumference AB, to E. then because D a. I. 3. A E E B b. 5. DA is equal to DB, the angle DAB is A equal to the angle DBA; and because * N. B. Whenever the expreffion "straight lines from the center" or “drawn ❝ from the center" occurs, it is to be understood that they are drawn tỏ the circumference. c. 16. I. d. 19. I. E Book III. AE a fide of the triangle 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 oppofite ; 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 straight 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 within it. Wherefore if any two points, &c. Q. E. D. 2. I. 3. b. 8. I. IF F a ftraight line drawn thro' the center of a circle, bifect a straight line in it which does not pafs thro' the center, it fhall 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 ftraight line drawn thro' the center bifect any straight line AB, which does not pafs thro' the center, in the point F. it cuts it also at right angles. Take a E the center 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 bafe EA is equal to the bafe EB; therefore the angle AFE is equal b to the angle BFE. but when a straight line standing upon another makes the adjacent angles equal to one another, each c.10.Def.1. of them is a right angle. therefore each of the angles AFE, BFE is a right angle; wherefore the straight line CD drawn thro' the center bifecting another AB that does not pass thro' the center, cuts the fame at right angles. d. 5. I. c E F B D But let CD cut AB at right angles; CD also bisects it, that is, AF is equal to FB. d The fame conftruction being made, because EA, EB from the center 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 are two an |