For, producing D'A to meet the circumference in E', and joining CE', the triangles ABD', AE' C, are similar, and give AB: AE' = AD': AC, whence AB × AC = AE' × AD': = (D'E' — D'A) × D'A = D'E' X D'A — D'A', or, by (58), AB X AC D'BX D'C-D'A'. = PROBLEMS OF CONSTRUCTION. PROPOSITION XXII.-PROBLEM. 68. To divide a given straight line into parts proportional to given straight lines. From A draw Let it be required to divide AB into parts proportional to M, N and P. an indefinite straight line AX, off AC M, CD = N, DE upon which lay = P, join EB, and draw CF, DG, parallel to EB; then AF, FG, GB, are proportional to M, N, P (18). A 69. Corollary. To divide a given straight line AB into any number of equal parts, draw an indefinite line AX, upon which lay off the same number of equal distances, each distance being of any convenient length; through M the last point of division on AX draw MB, and through the other points of division of AX draw parallels to MB, which will divide AB into the required number of equal parts. This follows both from the theory of proportional lines and from (I. 125). B PROPOSITION XXIII.-PROBLEM. MN P 70. To find a fourth proportional to three given straight lines. Let it be required to find a fourth proportional to M, N and P. Draw the indefinite lines AX, AY, making any angle with each other. Upon AX lay off AB = M, AD=N; and upon AY lay off AC P; join BC, and draw DE parallel to BC; then AE is the required fourth proportional. For, we have (15), = AB: AD= AC: AE, or M: N=P: AE. E Y 71. Corollary. If AB = M, and both AD and AC are made equal to N, AE will be a third proportional to M and N; for we shall have M: N-N: AE. PROPOSITION XXIV.-PROBLEM. 72. To find a mean proportional between two given straight lines. Let it be required to find a mean proportional between M and N. = Upon an indefinite line lay off AB = Second method. Take AB equal to the greater line M, and upon it lay off BC= N. Upon AB describe a semi-circumference, erect CD perpendicular to AB and join BD. Then BD is the required mean proportional (47). A ML D 73. Definition. When a given straight line is divided into two segments such that one of the segments is a mean proportional between the given line and the other segment, it is said to be divided in extreme and mean ratio. Thus AB is divided in extreme and mean ratio at C, if AB : AC = AC: CB. CI A C B = AC': C'B, If C' is taken in BA produced so that AB: AC' then AB is divided at C', externally, in extreme and mean ratio. PROPOSITION XXV.-PROBLEM. 74. To divide a given straight line in extreme and mean ratio. Let AB be the given straight line. At B erect the perpendicular BO equal to one half of AB. C' A B With the centre O and radius OB, describe a circumference, and through A and O draw AO cutting the circumference first in D and a second time in D'. Upon AB lay off AC AD, and upon BA produced lay off AC' = AD'. Then AB is divided at Cinternally, and at C'externally, in extreme and mean ratio. For, 1st, we have (59), = that is, AB is divided at C, internally, in extreme and mean ratio. 2d. The proportion [1] gives by composition (10), = or, since AD' AC', AD' + ABC'B, AB+ AD DD' + that is, AB is divided at C', externally, in extreme and mean ratio. 2 AC' AO+ 2 = AO' = AB2 + (4B)* = AB'. whence, extracting the square root, 5 AB 2 2 76. Definitions. When a straight line is divided internally and externally in the same ratio, it is said to be divided harmonically. Thus, AB is divided harmonically at C and D, if CA : CB = DA : DB; that is, if the ratio of the distances + C B of C from A and B is equal to the ratio of the distances of D from A and B. Since this proportion may also be written in the form the ratio of the distances of A from C and D is equal to the ratio of the distances of B from C and D; consequently the line CD is divided harmonically at A and B. The four points A, B, C, D, thus related, are called harmonic points, and A and B are called conjugate points, as also C and D. PROPOSITION XXVI.-PROBLEM. M NH 77. To divide a given straight line harmonically in a given ratio. Let it be required to divide AB harmonically in the ratio of M to N. Upon the indefinite line AX, lay off AE M, and from E lay off EF and EG, each equal to N; join FB, GB; and draw EC parallel to FB, ED parallel to GB. therefore, by the definition (76), AB is divided harmonically at C and D, and in the given ratio. 78. Scholium. If the extreme points A and D are given, and it is required to insert their conjugate harmonic points B and C, the harmonic ratio being given = M: N, we take on AX, as before, AE= M and EF EG = N, join ED, and draw GB parallel to ED, which determines B; then, join FB and draw EC parallel to FB, which determines C. = Also if, of four harmonic points A, B, C, D, any three are given, the fourth can be found. PROPOSITION XXVII.-PROBLEM. 79. To find the locus of all the points whose distances from two given points are in a given ratio. Let A and B be the given points, and let the given ratio be M: N. Suppose the problem solved, and that P is a point of the required locus. Divide AB internally at E Cand externally at D, in the ratio B M: N, and join PA, PB, PC, PD. PA: PB = = M: N= therefore, PC bisects the angle APB, and PD bisects the exterior angle BPE (23). But the bisectors PC and PD are perpendicular to each other (I. 25); therefore, the point P is the vertex of a right angle whose sides pass through the fixed points C and D, and the locus of P is the circumference of a circle described upon CD as a diameter (II. 59, 97). Hence, we derive the following Construction. Divide AB harmonically, at C and D, in the given ratio (77), and upon CD as a diameter describe a circumference. This circumference is the required locus. |