equal to the square of GH: But the squares of HE, EG are equal (1. 47.) 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, E F 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. Which was to be done. BOOK III. DEFINITIONS. I. EQUAL circles are those of which the diameters are equal, or 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 the centres are equal." II. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. III. Circles are said to touch one another, which meet, but do not cut one another. IV. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. V. And the straight line on which the greater perpendicular falls is said to be farther from the centre. VI. A segment of a circle is the figure contained by a straight line and the circumference it cuts off. VII. "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 contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. IX. And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. X. The sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them. XI. Similar segments of a circle are those in which the angles are equal, or which contain equal angles. 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 straight line AB, and bisect (1. 10.) it in D; from the point D draw (I. 11.) 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 the centre be in CE, it must be in the middle point of CE, that is, it must be in F: C But if it be not in CE, let, if possible, & be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre G*; therefore the angle ADG is equal (1. 8.) to the angle GDB: But when F E G D B a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (1. Def. 10.): 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, *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. which is impossible: Therefore G is not the centre of the circle ABC: In the same 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. 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. If any two points be taken in the circumference of a circle, the straight line which joins them shall 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 shall fall within the circle. For, if it do not, let it fall, if possible, without, as AEB; find (III. 1.) D, the centre of the circle ABC, and join AD, DB; in the circumference AB A C F E B take any point F, join DF, and produce it to E: Then, because DA is equal to DB, the angle DAB is equal (1. 5.) to the angle DBA; and because A E, a side of the triangle DAF, is produced to B, the angle DEB is greater (1. 16.) than the angle DAE; but DAE is equal to the angle DBE; therefore the angle DE B is greater than the angle DBE; but to the greater angle the greater side is opposite (1. 19.); 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. PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect 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 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 cuts it also at right angles. Take (III. 1.) 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 (1. 8.) to the angle BFE But when a straight line standing upon another makes the adjacent angles equal to one A another, each of them is a right E F B D (1. Def. 10.) angle: Therefore each of the angles AFE, BFE is a right angle; wherefore the straight line CD, drawn through the centre bisecting another AB that does not pass through the centre, cuts the same at right angles. But let CD cut AB at right angles; CD also bisects it, that is, AF is equal to FB. The same construction being made, because EA, EB from the centre are equal to one another, the angle EAF is equal (1. 5.) to the angle ECF; and the right angle AFE is equal to the right angle BFE: Therefore in the two triangles EAF, EBF, there are two angles in one equal to two angles in the other, |