Proposition 34. Problem. 347. To find a fourth proportional to three given straight lines. Given, the three lines M, N, P. Required, to find a fourth propor tional to M, N, P. Cons. Draw the two indefinite straight lines AG, AH, making any with each Join DB, and through C draw CE || to DB. Then DE is the fourth proportional required. Proof. Since BD and CE are ||, .. AB: BC = AD: DE. M N P H E A st. 1. || to a side of a ▲ divides the other two sides proportionally (298). But ABM, BCN, AD = P. .. M: NP: DE. (Cons.) Q. E.F. 348. SCH. If the lines N and P are equal, then BC and AD are both laid off equal to N, and DE is the third proportional to M and N (280). The proportion in (347) then becomes M:NN: DE. EXERCISES. 1. Construct a fourth proportional to the lines 3, 7, 11. 2. If from D, one of the angles of a parallelogram ABCD, a straight line is drawn meeting AB at E and CB produced at F; show that CF is a fourth proportional to EA, AD, and AB. Proposition 35. Problem. 349. To find a mean proportional between two given straight lines. 350. SCH. The mean proportional between two lines is often called the geometric mean, while their half sum is called the arithmetic mean. EXERCISE. From a given point P in a circle a perpendicular PM is drawn to a given chord AB; from A, B perpendiculars AC, BD are drawn to the tangent at P: prove that PM is a mean proportional between AC and BD. 351. DEF. A straight line is said to be divided in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. Thus, the line AB is divided in extreme and mean ratio at C if A C B AB AC = AC: CB. Proposition 36. Problem. 352. To divide a given straight line in extreme and mean ratio. Join AC, cutting the O at D, and produce it to meet Then F is the pt. of division required. Proof. Since AB is a tangent and AE a secant to the O, ... AE: AB: = AB: AD. (1) The tangent is a mean proportional between the whole secant and the external segment (339). .. by division, AE - AB : AB = AB - AD: AD. AB DE and AD AF. But .. AE. = = ABAD = AF, and AB (288) (Cons.) AD = FB. Substituting these values in the above proportion, we have AF: AB = FB: AF. (285) .. AB: AF = AF: FB. ... AB is divided in extreme and mean ratio at F. = Q. E. F. 353. SCH. If BA be produced to the left of A to a point F' so that AF' AE, then F' is another point of division having the same property as F. Thus we have from proportion (1) in (352), by composition, AE AB AE = ABAD: AB. (287) AF'. (Cons.) But AE AB BF', and AB+ AD or = .·. BF' : AF′ = AF′ : AB, AB: AF' = AF': F'B. ... AB is divided in extreme and mean ratio at F'. AB is said to be divided at F internally, and at F' externally, in extreme and mean ratio. NOTE. This division of the straight line was called by the ancient geometers the golden section. Proposition 37. Problem. 354. On a given straight line to construct a polygon similar to a given polygon. Then A'B'C'D'E' is the required polygon. Proof. Since the As ABC, A'B'C' are mutually equi ... the polygons ABCDE, A'B'C'D'E' are similar, being composed of the same number of As similar each to each, and similarly placed (320). Q.E.F. APPLICATIONS. 1. To find the altitudes of a triangle in terms of its sides.* See (332). Let ABC be a A; denote by a, b, c, the lengths of the sides opposite the Zs A, B, C, respectively, and by h the altitude AD. Of the two s B and C, at least one is acute; suppose it to be the /C. Then: ▲ ADC, h2 = b2 — CD3. In the B (328) ... CD = 2a 4a2b2-(a2+b2 — c2)2 (2ab+a2+b2—c3) (2ab—a2 —b2+c2) [(a+b)2 —c2] [c2 — (a—b)2] 4a2 (a+b+c)(a+b−c)(c+a−b)(c−a+b) 4a2 a+b+c= 28. a+b-c=2(s — c). cab=2(s - b). 2(s—b). ca+b=2(sa). (1) Substituting in (1) and extracting the square root, we 2. To find the medians of a triangle in terms of its sides, See (333). * See Traité de Géométrie, par Rouché et Comberousse, p. 139. + The difference of the squares of two numbers equals the product of the sum and difference of the two numbers, |