THEOREMS ON CHAPTER XIV. THEOREM LXXXVI. If a secant is drawn through a fixed point in the plane of a circle, the rectangle contained by the distances of that point from the two points of intersection with the circumference is constant, and conversely. Let two secants be drawn from the fixed point E, the first cutting the circle in the points A and B, and the second in C and D. Then shall Join the points A D, C B, then in the triangles A E D, CEBLADE4 CBE because they are angies at the circumference on the same arc. Also the angles at E are equal. Hence, the triangles are similar and consequently, Conversely, If two straight lines A B and C D, produced if necessary, meet in a point, E, so that the extremities A, B, C, D, must be situated on the saine circumference. For the triangles A E D and CEB, having the sides about the equal or coincident angles at E proportional, must be similar. Consequently the homologous angles DAE, BCE are equal, and therefore the points A, B, D, and C lie on the same circumference. THEOREM LXXXVII. If from a point without a circle a tangent and a secant be drawn to it, the square on the tangent will be equal to the rectangle contained by the whole secant and the part of it without the circle, and conversely. Let EF be a tangent and EDC a secant from E, then shall the square on EF be equal to the rectangle ED EC. Draw the diameter F B; join E B, and let A be the point in which it cuts the circumference, and join F A. Because EFB is the angle between a diameter and tangent it is a B right angle. Because FA B is the angle of a A semicircle, therefore F A is perpendicular to E B. E Hence the square on EF is equal to the rectangle EA EB (Theorem LI.), and therefore also to the rectangle ED EC. Conversely, if two points, A, B, in one side of an angle, E, and a third point, F, in the other side, be so situated that the square on E F the rectangle EAE B, the circumference which passes through all three points will touch the side E F at the point F. = In this case the triangles E B F and A FE are similar, because they have the sides about the common angle E proportional; hence the angles AFE, EBF are equal; and if a circumference be described, passing through A, B, and F, since E F makes with AF an angle equal to the angle in the opposite segment, E F will touch the circle. THEOREM LXXXVIII. The side of a regular decagon inscribed in a circle is the greater segment when the radius is divided in extreme and mean radius. Let ABCDEFG be an inscribed regular decagon. AF, BG, AD, B D. Then BAD and ▲ D A F are angles at the circumference on two divisions of the circumference, and ZAOB is an angle at the centre on one; therefore LBAM = <MAO=4AOM. Hence the triangles A M B, OA B have the angles of one respectively equal to the angles of the other, and since A OB is isosceles, so also is AM B; K B C E G F therefore A B=AM=MO. Because A M bisects OA B, OA: AB=OM: MB; D that is OB:0M=OM: M B. OB is divided in extreme and mean ratio. THEOREMS FOR EXERCISE. 117. If two circles intersect, the common chord produced bisects the common tangent. 118. The tangents drawn from a point to the same circumference are equal. 119. If two diagonals of a regular pentagon be drawn to cut one another, they will be divided in extreme and mean ratio, and the greater segment will be equal to the side of the pentagon. 120. If two equal chords of a circle cut one another either within or without a circle, the segments of the one between the point of intersection and the circumference shall be equal to the segments of the other, each to each. 121. If tangents be drawn at the extremities of any two diameters of a circle, and produced to intersect one another, the straight lines joining the opposite points of intersection will both pass through the centre. 122. If any chord of a circle be produced equally both ways and tangents to the circle be drawn on opposite sides of it from its extremities, the line joining the points CHAPTER XV. MAXIMA AND MINIMA. Definitions. Among quantities of the same kind, that which is greatest is called a maximum, and that which is smallest a minimum. Thus the diameter of a circle is a maximum among all inscribed straight lines; and a perpendicular is a minimum among all the straight lines drawn from a given point to a given straight line. Applying these terms, Theorem L. may be stated thus:-" The maximum of rectangles formed by the segments of a given straight line occurs when the segments are equal, and any other differs from the maximum by the square on the line which is half the difference of the segments." Several other cases have occurred among the preceding propositions in which the hypotheses of a theorem or the conditions of a problem are restricted within certain limits as to magnitude and position, which constitute maxima or minima. Thus, the limit to which the sum of two sides of a triangle on a given base can be reduced is the base itself (Theorem XII.); the greatest line which can be drawn from a given point within a circle to the circumference is the line which passes through the centre; and the least line which can be so drawn from the same point is the part or part produced of the greatest line between the given point and the circumference. These examples will be extended and other cases added in the present chapter. For example, it will be proved that (a) The minimum of all lines between the same two points is a straight line; (b) If polygons having the same perimeter are compared, it is found that-(1) Of those having the same number of sides the greatest is regular-(2) Of two regular polygons, the greater is that which has the greater number of sides; (c) The perimeter of a polygon inscribed in a circle is less than the circumference, and the perimeter of the circumscribing circle is greater than the circumference; (d) Since the perimeter of the circumscribing square is four times the diameter, and that of the inscribed hexagon three times the diameter, it follows that the circumference is between three and four times the diameter. Ratio of the Circumference of a Circle to the Diameter. From proposition (c) we are able to find approximately the ratio of the circumference to the diameter. Consider two regular polygons of the same number of sides inscribed in two circumferences; the ratio of the perimeter of the polygons is equal to the ratio of the radii of the circumferences. Hence the ratio of the perimeter of a regular polygon to the radius of the circle circumscribing it is the same in both figures. This is true, whatever may be the number of sides in the polygon; but as the number of sides increases, the perimeter of the polygon approaches more and more nearly the circumference of the circle circumscribing it. We conclude, therefore, that the ratio of the circumference of a circle to its diameter is always the same. The constant value of this ratio cannot be expressed exactly by any finite number. We will represent it by the Greek letter (pronounced pi). It may be found approximately by taking the ratio of the perimeter of an inscribed or circumscribed regular polygon to the diameter. For example, the side A B of a regular hexagon inscribed in a circle is equal to the radius; hence the perimeter is equal to three times the diameter (Fig. 122), and the perimeter of the circumscribed hexagon 3.4641 times the diameter; hence the ratio expressed by lies between 3·0 and 3·4641. By taking a polygon with a greater number of sides we get a closer approximation; thus, a polygon of 128 sides shows us that lies between or 318 and 31. A nearer approximation still is 3·1416, 7 22 or 3.14159, The fraction for, written thus = it consists of the first three odd numbers taken twice, with a bar in the middle. To draw a Straight Line which shall have the Length of the Circumference of a given Circle. It is impossible to solve this problem exactly, but a line may be drawn which differs from the circumference by a very small fraction of its length, by adding to three times the diameter the fifth part of the side of the inscribed square. The construction is indicated in the figure (Fig 188). The side, A C, of the inscribed square= = the radius MA FIG. 188. N X2; hence A M, the fifth of A C, = 1414 of the diameter; hence, |