V. The angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the base of the segment, may be called the angle in a segment. VI. An angle may be said to stand on the circumference intercepted between the straight lines which contain the angle. VII. When straight lines are said to be drawn from the centre, it is to be understood that 1 they are drawn to the circumference. VIII. The figure contained by two straight lines drawn from the centre, and the circumference between them, may be called a sector of a circle. PROPOSITION I. THEOREM. If the radius of one circle is equal to the radius of another circle, the circles are also equal. Let ABG, DEF be two circles, having the radius AC of the one circle equal to the radius DI of the other circle, the circles will be equal. For, if the circle ABG be applied to the circle DEF, so that the centre C of the one, shall be upon the centre I of the other; then, A B D G F because AC is equal to DI, and because all the straight lines from the centre to the circumference are equal (Def. 54. 1.), the circumference of the circle ABG will fall upon the circumference of the circle DEF, and the circle ABG will be equal to the circle DEF. Therefore, if the radius of one circle is equal to the radius of another circle, the circles are also equal. Which was to be proved. PROP. II. THEOR. The diameter divides the circle into two equivalent parts. Let ABCD be a circle, and BD the diameter; DAB is equivalent to BCD. For, if DAB be applied to BCD, because all the lines from the centre to the circumference are equal (Def. 54. 1.), the circumference DAB will fall upon the circumference BCD, and the A segment DAB will be equivalent to the segment BCD. Therefore, the diameter divides the circle into two equivalent parts. Which was to be proved. B D PROP. III. PROB. To find the centre of a given circle. C Let ABC be the given circle; it is required to find its centre. Draw any straight line AB in the circle ABC, and bisect (8. 1.) it in D; from the point D draw (9. 1.) DC at right angles to AB, and produce it to E, and bisect (8. 1.) 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 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 sides BD, DG, each to each; and the base AG is equal to the base BG, because they are drawn from the centre to the circumference (Def. 54. 1.); therefore the angle ADG is equal (7. 1.) to the angle BDG; but when a straight line standing upon another straight line, makes the adjacent angles equal to one another, each of the angles is a right angle (Def. 27. 1.); therefore BDG is a right angle; but BDF is likewise a right angle; wherefore the angle BDF is equal to the angle BDG, the greater to the less, which is impossible; therefore G is not the centre of the circle ABC. In the same manner it can be shewn, that no other point but F is the centre; that is, F is the centre of the circle ABC. Which was to be found. F G A. B D E COR. If a straight line bisect another straight line in the circle at right angles, the centre of the circle is in the line which bisects the other. PROP. IV. THEOR. If two points be taken in the circumference of a circle, the straight line which joins them shall fall wholly within the circle. Let ABC be a circle, A and B any points in the circumference; straight line AB is wholly within the circle. Find the centre of the circle D (3. 3.), draw DC perpendicular to AB (10. 1.), and join DA, DB; and draw DF any straight line from Dupon AB; DF is less than DA. Because DF is nearer the perpendicular DE, it is less than DA (13. 1.): in the same way it may be shewn that any other straight line from D to AB is less than DA; and therefore DF to meet the circumference, must D A B F F the be produced beyond AB. Therefore the circumference is beyond AB, or AB is wholly within the circle. Q. E. D. PROP. V. 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 (3. 3.) 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 A E D F 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 (5. 1.) 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 angles in one equal to two angles in the other, and the side EF which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal (6. 2.); AF therefore is equal to FB. Wherefore, if a straight line, &c. Q. E. D. PROP. VI. THEOR. Equal straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to one another. Let the straight lines AB, CD in the circle ABDC be equal to one another; they are equally distant from the centre. Take E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD; then, because the straight line EF passing through the centre cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects (5: 3.) it; where A E G fore AF is equal to FB, and AB double of AF; for the same reason CD is double of CG, and AB is equal to CD, therefore AF is equal to CG; and because AE is equal to EC, the square of AE is equal to the square of EC; but the squares of AF, FE are equal (17. 2.) to the square of AE, because the angle AFE is a right angle; and for the like reason the squares of EG, GC are equal to the square of EC; therefore the squares of AF, FE are equal to the squares of CG, GE, of which the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line FE is therefore equal to EG; but straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal (Def. 2. 3.); therefore AB, CD are equally distant from the centre. B D Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG; AB is equal to CD; for, the same construction being made, it may, as before, be demonstrated that AB is double of AF, and CD, double of CG, and that the squares of EF, FA are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG, and AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore, equal straight lines, &c. Q. E. D. PROP. VII. THEOR. The diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the less. Let ABCD be a circle, of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG; AD is greater than any straight line BC which is not a diameter, and BC greater than FG. From the centre draw EH, EK perpendicular to BC, FG, and join EB, EC, EF; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC; but EB, EC F are greater (Def. 16. 1.) than BC; wherefore, also, AD is greater than BC. And, because BC is nearer to the centre than FG, EH is less (Def. 3. 3.) than EK; but, as it was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal K B H E |