PROP. XIV. PROB. In the same straight line, three points being given, to find a fourth, so that the rectangle under its distance from the first and a given line, shall have a given ratio to the rectangle under its distances from the second and third points. Let it be required to find a point D, such that AD×G: CD x BD :: M: N. = CD x BD; wherefore, by the last proposition, the point of section D is given. COMPOSITION. Having made G: H:: MN, find, by the last proposition, the point D, so that CD x BD = AD x H; Dis the section required. For (V. 24. cor. 2. El.) AD × G : AD × H, or CD.BD :: G : H, or M: N. PROP. XV. PROB. In the same straight line, three points being given, to find a fourth, so that the square of its distance from the first, shall be equal to the rectangle under its distances from the second and third points. Let it be required to find a point D, such that AD1= CD.BD. 1. When the point D lies between A and B. ANALYSIS. Because AD' = CD.BD, it follows (V. 6. El.) that CD: AD: AD: BD; whence (V. 9. El.) AC: AD :: AB: BD, and al the ratio of AC: AB being given, the ratio of AD: BD is given; and since AB is given, the point D (VI. 4. El.) is given. COMPOSITION. Divide AB (VI. 4. El.) in the ratio of AC to AB, and the point of section D is that required. For, because AD: BD:: AC: AB, AD: BD::AC-AD, or CD: AB-BD, or AD (V. 19. cor. 1. El.); whence (V. 6. El.) AD2 = BD.CD. 2. When the point D lies between B and C. ANALYSIS. Make DEAD, and since AD CD.BD, CD: AD, or DE:: AD: BD, and therefore (V. 10. El.) CE: DE :: AB: BD, and alternately CE: AB:: DE: BD, or (V. 3. El.) CE: AB:: 2DE, or AE: 2BD; whence (V. 9. El.) CE:AB::CE+AE, or AC AB + 2BD, or BE, and consequently (V. 6. El.) CE.BE AB.AC; but the rectangle AB.AC being given, the rectangle CE.BE is likewise given, and BC being given, the point E is given (VI. 20. El.), and therefore D, the bisection of AE, is given. COMPOSITION. Divide BC (VI. 20. El.) in E, such that CE.BE AB.AC, and bisect AE in D; then AD' = CD.BD, For since CE.BE AB.AC, it is evident that CE: AB :: AC: BE; whence (V.S El.) CE: AB;: AE: 2BD, or DE: BD; and alternately, CE: DE :: AB : BD, and (V. 9. El.) CD: DE, or AD :: AD : BD; wherefore (V. 6. El.) CD.BD AD'. This last case is evidently subject to limitation; for the rectangle AB.AC being equal by construction to CE.BE, must not exceed the square of the half of BC, which (II. 23. cor. 1. El.) is the greatest rectangle contained under the segments of BC. Consequently, if E coincide with the middle point O, it limits the problem; but then AB. AC=BO3, or AB.AC + BO2 = (II. 23. cor. 2. El.) AO2 = 2BO3, and therefore AO is the diagonal of a square described on BO. Whence AB: BC:: 2-1: 2, or i: 28; that is, the ratio of AB to BC has attained its maximum, when it is that of half the side of a square to the sum of the side and the diagonal. PROP. XVI. PROB. In the same straight line, three points being given, to find a fourth, such that the square of its distance from the first, shall have a given ratio to the rectangle under its distances from the second and third points. Let it be required to find a point D, such that AD' shall be to CDX DB in a given ratio. 1. When D lies between the points A and B. ANALYSIS. On BC describe a semicircle, and draw the tangent DE; then (III. 36. El.) DE = CD x DB, and consequently the square of AD is to the square of DE in the given ratio; whence the ratio of AD to DE is given. Join EF, and produce ED to meet AG a perpendicular to AC. The triangles ADG and EDF are evidently similar, and therefore AD:AG::DE: EF, or alternately AD: DE :: : AG EF; and since the ratio AD to DE is given, the ratio of AG to EF is also given, and the radius EF being given, AG and the point G are thence given; wherefore the tangent GE and its intersection D with AC, are given. COMPOSITION. Let M: N be the given ratio, and to these find (VI. 18. El.) a mean proportional O, on BC describe a semicircle, make O: M:: BF: AG, a perpendicular erected from A, and (III. 30. El.) draw the tangent GDE; the intersection D is the point required. For, the triangles DAG, and DEF being similar, AD:AG:: DE: EF, and alternately AD: DE:: AG: EF, or M: 0; wherefore (V. 21. cor. 1. El.) AD: DE :: M3: 0, that is, (V. 23. El.) M: N; but (III, 36. El.) DE2=CD× DB, and consequently AD1: CD× DB:: M: N. 2. When D lies between the points B and C. ANALYSIS. On BC describe a semicircle, draw DF perpendicular to the diameter, and meeting the circumference in F, and join AF. Because (III. 36.) BD × DC=DF, the ratio of AD' to DF is given, and consequently that of AD to DF; but fore the intersection For F', the perpendicular FD, or F'D', and the point D, or D', are all given. COMPOSITION. Let M:N express the given ratio, and to these find (V. 18. El.) a mean proportional O, make (VI. 3. El.) M to O as AC to the perpendicular CE, join AE meeting the circumference of a semicircle described on BC in the point F or F, and let fall the perpendicular FD or F'D'; then M: N: AD: BDx DC, or AD: BD x D'C. For the triangle ACE is evidently similar to ADF or AD'F', and therefore AC : CE :: AD: DF, and ACa : CE :: AÐ2: DF1; but (V. 23. El.) M : N :: M2 : O3, or as AC: CE, and consequently AD: DF, that is, BD × DC :: M: N.. This problem evidently requires limitation; for, if AE should diverge too much from AC, it will not meet the circumference at all. Hence an extreme case will occur, when AE touches the circle. But the ratio of AC to CE, or of AD to DF, will then be the same as that of a tangent from A is to the radius HB; and consequently the limiting ratio is the duplicate of this,—or the ratio of M to N can never approach nearer to the ratio of equality than that of AB X AC, or AH'— HB3, to HB2. |