CHAPTER VII. 85. THEOREM. If two chords intersect within a circle, the rectangle on the segments of one chord equals the rectangle on the segments of the other. Let AB and CD represent two chords intersecting at I. Draw the auxiliary lines AC and DB. The vertical angles at I are equal; Z CAI 2BDI, each being subtended by the arc CKB. ZACI=2DBI, each being subtended by the arc AD. .. AAIC is similar to ▲ DIB. K FIG. 133. B Then = ID AI CI Q. E. D. Exercise. х If - Show how to construct a square that shall be equivalent to a given rectangle, the sides of which are a and b. two chords be so drawn that the segments of one were adjacent sides of a rectangle and the segments of the other were equal, by the theorem we would have a rectangle equivalent to a square. We know (§ 45) that if a diameter be drawn perpendicular to a chord it will bisect the chord. Hence if a circumference be constructed with bpa FIG. 134. (a + b) as a diameter, and at the common extremity of a and b a perpendicular chord be constructed, half the chord will be the side of the required square, since ab = x2. The equation x2 ab is thus solved geometrically and exactly. 86. THEOREM. If two secants intersect without the circumference, the rectangle on the distances from the common point to the two intersections with the circumference in one case will be equal to the rectangle similarly formed in the other. The two ADAC and BEC are similar. C is common. E Exercises. — 1. Having given a rectangle, construct an equivalent rectangle that shall have a given side. 2. If one secant remain stationary, and the other rotate about the common point C until it become a tangent, we shall have a secant and a tangent. Show that CT2 = CAX CB. T 3. Use Exercise 2 to construct a square equivalent to a given rectangle. 4. Construct a square equivalent to a given triangle. FIG. 136. 87. PROBLEM. Find an expression for the bisector of an angle of a triangle in terms of the including sides and the segments of the third side. Circumscribe a circle about the triangle, and draw the auxiliary line AK. AKAD and If it be desired to find m and n in terms of the sides a, c, and d of the triangle, The solution of this pair of equations will give m and n, in terms of a, c, and d. Q. E. F. 88. PROBLEM. To find a relation between the sides and the diagonals of a quadrangle inscribed in a circle. Let a, b, e, and d represent the sides of the inscribed quadrangle, and m, n, p, and q the segments of the diagonals. The Ambq and dpn are similar (being mutually equiangular). a n b m FIG. 138. (1) Because the eng and apm are similar (being mutually Adding the members of equations (1) and (2), we have Equation (3) expresses a relation between the sides and segments of the diagonals, but a more convenient relation may be obtained by transforming the second member. By $$ 75 and 76, representing the perpendicular projection of p on the other diagonal by h, d2 = p2 + n2 ± 2 nh, a2 = p2 + m2 = 2 mh. (4) (5) Multiplying both members of (4) by m, and both members of (5) by n, we have But md2 = mp2 + mn2 ± 2 mnh, na2 = np2 + m2n = 2 mnh, md2 + na2 = (m + n) p2 + (m + n) mn. mn = pq. ... md2 + na2 = (m + n) p2 + (m + n) pq (§ 85) · = (m + n) (p + q). The sum of the rectangles on the opposite sides equals the rectangle on the diagonals. Q. E. F. 89. PROBLEM. Show how to construct a square that shall have the same ratio to a given square as two given Analysis. If the required square were known, and if sides of the given and required squares were placed at right angles to each other and so that they had a common extremity, by joining their other extremities a right triangle would be formed; and if from the vertex of the right angle a perpendicular were drawn to the hypothenuse, it would separate the latter into segments proportional to the squares on the corresponding sides. [§ 77 (c).] Construction. Remembering that if a circumference be constructed with the hypothenuse of a right triangle as its diameter, it will pass through the vertex, we have, taking m and n of convenient length and so that erecting a perpendicular at their common extrem m n = h k If a side of the given square be less than a, lay off VQ |