PROP. XII. PROB. Through two given points, to describe a circle bisecting the circumference of a given circle. Let A and B be two points, through which it is required to describe a circle ADGEB, that shall bisect the circumference of the circle HDFE. ANALYSIS. Join D, E, the points of intersection. Because DFE G. Wherefore the three points A, G, and B being given, the circle AGB is (III. 11. El.) given. COMPOSITION. Through C, the centre of the given circle, draw ACF, make (VI. 3. El.) AC: HC :: CF, or HC: CG, and through the three points A, G, and B, describe (III. 11. cor. El.) the circle AGB: This will bisect the circumference HDFE. For, through one of the points of intersection, draw the diameter DCI, and produce it to meet the circumference S of the circle AGB in K. Because AC: HC :: HC:CG, the square of HC is (V. 6. El.) equal to the rectangle AC, CG; but (III. 36. El.) HC = DC.CI, and AC.CG = DC.CK; wherefore DC.CI = DC.CK, and CI = CK, or the points I and K are one, and the circle AGB passes through both extremities of the diameter of HDFE. PROP. XIII. PROB. To find a point in the diameter of a circle, such that the square of a straight line inflected from it at a given angle to the circumference, shall have a given ratio to the rectangle under the segments of the diameter. Let it be required to draw DE at a given angle with DB, and so that the square of DE shall have a given ratio to the rectangle AD, DB. ANALYSIS. Make EG = FD, join CF, draw the radius CGH, join CEG equal to CFD, and the sides F CE and EG equal to CF and FD, are equal, and consequently the angle ECG is equal to FCD; whence (III. 15. El.) the arc HE is equal to AF, and therefore (III. 22. cor. El.) AH is parallel to DE. But the angle BDE is given, and thence BAH; wherefore the chord AH is given. Again, the rectangle AD,DB, being equal to FD,DE (III. 36. El.), is also equal to DE, EG; and therefore DE* is to DE.EG, or (V. 24. cor. 2. El.) DE is to EG, in the given ratio; but (VI. 2. El.), DE: EG :: AI: IH, consequently Al is to IH in a given ratio, and hence AH is to HI in a given ratio. Wherefore, since AH is given, IH and the point I are given; and thence IC, the point E, and DE, are all given. COMPOSITION. Draw AH at an inclination with AB equal to the given angle, and produce it to I, so that AI shall be to AH in the given ratio, join IC, and draw ED parallel to IA; D is the point required. Because AI: IH:: DE: EG, DE is to EG in the given ratio, and consequently DE* is to DE.EG in the same ratio. But FE being parallel to AH, the arc HE is equal to AF, and thence the angle HCE is equal to ACF; the triangles CGE and CDF, having thus the side CE equal to CF, and the angles ECG and CEG equal to FCD and CFD,-are equal, and hence the side EG is equal to FD. Wherefore DE.EG = DE.FD = AD.DB, and consequently DE is to AD.DB in the given ratio. PROP. XIV. PROB. Through two given points, to draw straight lines to a point in the circumference of a given circle, so that the chord of the intercepted segment shall be parallel to the straight line that connects the given points. Let it be required, from the points A and B, to inflect AC and BC cutting the given circumference in D and E, such that DE shall be parallel to AB. ANALYSIS. Draw the tangent DF meeting AB in F. The angle given, the rectangle AD, AC is also A 13 D E given; for it is equal to the square of the tangent AG (III. 36. cor. 2. El.), when A lies without the circumference, and equal to the square of AG (III. 36. cor. 1. El.) a perpendicular to the diameter, in the case where that point lies within the circle. Hence the rectangle AF, AB is given; and AB being given, AF is likewise given, and consequently the point F. Wherefore the tangent FD is given in position; and since the point A is given, the straight line AC is given, and thence BC and the intersection E. COMPOSITION. If the point A be without the circle, draw the tangent AG; or if it lie within the circle, erect AG perpendicular to the diameter which passes through it. Make (VI. 3. El.) AB: AG :: AG: AF, from F draw the tangent FD, join AD, and produce it to meet the opposite circumference in C, join CB, cutting the circle in E; the straight line DE is parallel to AB. For, since AB: AG :: AG: AF, AG = AB.AF; but (III. 36. cor. 1. and 2. El.) AG = CA.AD, whence AB.AF = CA.AD, and consequently (V. 6. El.) AB : AC :: AF: AD. Wherefore (V. 15. El.) the triangles BAC and DAF, having the sides about their common angle proportional, are similar, and hence the angle ACB is equal to AFD; but (III. 29. El.) ACB or DCE is equal to EDF, and consequently the angle AFD is equal to EDF, and (III. 22. cor. El.) the chord DE is parallel to AB. PROP. XV. PROB. From two given points in the circumference of a given circle, to inflect straight lines to another point in the opposite circumference, such as to intercept, on either side of the centre, equal segments of a given diameter. Let it be required, from the points A and B, to inflect AC and BC, so as to intercept, on the diameter DE, equal portions from the centre. ANALYSIS. Join BA, and produce it and the diameter ED to meet in M, draw COL, from O let fall the perpendicular OK upon AB, join LK, through A draw AHI parallel to DE, and join HK. |