greater segment as x, then the less segment is (1-x), and the following quadratic equation is obtained, x2= 1−x, or x2+x=1, √5-1 and therefore x = minate mixed decimal, and therefore diminishing it by 1, and dividing the remainder by 2, gives also an interminate decimal. By means of the corollary, however, a number of interesting approximations may be made to the numerical solution; for if the whole line be first considered to be 2, and its segments each 1, then the line made up of the whole and its greater segment (Cor.) will be 3, the greater segment 2, and the less segment 1; again, repeating the same operation with 3 and its greater segment, the number 5 is obtained, its greater segment being 3, and its less segment 2; and so on to any extent, by which the following interesting approximations are obtained: But the square root of 5 is an inter 2) the product is greater than the square, by the whole square. 1 f 4) the product is less than the square, by a fourth part of the square. 10 the product is greater than the square, by a ninth part of the square. 9 24 25 65 64 168 169 the product is less than the square, by a twenty-fifth part of the square. the product is greater than the square, by a sixty-fourth part of the square. the product is less than the square, by a hundred and sixty-ninth part of the square. 441) the product is greater than the square, by th part of the square. 55 × 21 = 1155 the product is less than the square, by 34 × 34 = 1156 11th part of the square. It is evident, from the above, that the product is alternately greater and less than the square; and that it differs from the square by a less part of the square, the greater the number of parts into which the line is divided. When the line was divided into two parts, the product was double of the square; but when it is divided into 55 parts, and the greater taken 34 and the less 21, the product is then less than the square by Tth part of the square. In the same manner, the difference may be made much less, by taking a greater number of divisions in the line; the next number of divisions would be 89, and then the product would exceed the square byth part of the square. In the demonstration of the second part of this proposition, the following is assumed as an axiom: 'If there be four quantities such that the first is double of the second, and the third also double of the fourth, the difference of the first and third will also be double of the difference of the second and fourth.' This not having been stated as an axiom, nor being sufficiently self-evident, the following demonstration may be substituted. оя MT Let D, the centre of the circle, be without the angle BAC; join AD, and produce it to E, and let AB, DC intersect each other in O. The exterior angle BOC of the triangle BOD is equal to the angles ODB and DBO; but the angle DBQ is equal to the angle DAO, because DB is equal to DA, therefore the angle BOC is equal to the angles ODB and DAO. Again, the exterior angle BOC of the triangle AOC is equal to the angles OCA and OAC; and since DA is equal to DC, the angle OCA is equal to the angle DAC, which is equal to the two angles CAO and DAO, therefore the angle BOC is equal to twice the angle OAC and the angle DAO. But the same E B angle was proved to be equal to the angles ODB and DAO, therefore the angles ODB and DAO are equal to twice the angle CAO and the angle DAO; take away the common angle DAO, and there remains the angle ODB equal to twice the angle CAO; that is, the angle at the centre is double of the angle at the circumference. If two chords in a circle cut one another, the rectangle con tained by the segments of one of them is equal to the rectangle u contained by the segments of the other. Let the two chords AC, BD, within the circle ABCD, cut one another in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by DEEB. Join FA, FB, FC, and FD, F being the centre. Then in the isosceles triangle AFC, the square on AF is equal to the rectangle AE EC and the square on FE (II. D). Again, in the isosceles triangle DFB, that is, of AF, square on FE (II. D). D A F E the square on DF, is equal to the rectangle DE EB and the Therefore the rectangle AE⚫ EC and the square on FE is equal to the rectangle DEEB and the square on FE; take the square on FE from each, and there remains the rectangle AE⚫EC equal to the rectangle DE EB. In the same manner, as stated in the Appendix to the Second Book, this proposition may be applied practically to find the segments of lines drawn in a circle and cutting each other, for when three of the segments are given, the fourth may be found. The following exercises are appended to shew the way in which it may be applied. The letters refer to the figures in the text of Book Third. (1.) The chord of an arc is 24, and its height 8, find the diameter of the circle. Referring to figure (2) AC = 24, and BE = 8; therefore AE and EC are each 12; hence, since BE · ED = AE⚫EC, 8 × DE = 12 × 12 = 144; hence DE= 18. But BE + DE = the diameter 8 +18= 26. (2.) A segment of a wheel being found, its chord was measured, and found to be 18 inches, and the height was found to be 3 inches; what was the diameter of the wheel when entire ? Ans. 30 inches. (3.) The diameter of a circle is 36 inches, and the height of an arc is 12 inches; find the chord of the arc. Ans. 24√2 inches. (4.) Two chords cut one another within a circle, the segments of the one are 8 and 9, one of the segments of the other is 6; find the other segment, and which of them cuts off the greater segment from the circle. Ans. 12; and that whose segments are 6 and 12 cuts off the greater segment, for this chord is 18, and the other 17, it is therefore nearer to the centre (III. 15). PROPOSITION XXXVI. (1.) If from a point without a circle, whose diameter is 12, a straight line is drawn through the centre, whose whole length is 16; what is the length of a tangent to the circle drawn from the same point? Here since the whole line passing through the centre is 16, and the diameter is 12, the part without the circle is 4; hence, from the proposition, the square of the tangent is equal to 16 × 4, and therefore the tangent: 16 x 4 = √64 = 8. (2.) If the length of a tangent to a circle drawn from a point without it be 12, and the length of a line passing through the centre and drawn from the same point be 18, find the distance of the point from the circle and the diameter of the circle. = Referring to figure (1), by the proposition the rectangle AD DC is equal to the square on BD, or AD × DC = BD2, but AD = 18, and BD = 12; .. 18 x DC = 122, hence DC 1448. Again, AD = AC + CD, or AC AD - CD = 18-810; the distance of the point without the circle is therefore 8, and the diameter of the circle is 10. = (3.) Find the diameter of the earth, having given that from the top of a hill one mile high, an observer may see to the distance of 89 miles. Ans. 7920. (4.) Having given that the earth is a sphere whose diameter is 7920 miles, find how far above its surface the eye of an observer must be placed to see to a distance of 10 miles. Ans. 663 feet nearly. (5.) How far must a point be from the circumference of a circle whose diameter is 12, so that the tangent drawn from it to the circle may be 8? Ans. 4. (6.) Find, in terms of the diameter, the distance of a point without a circle, from which a tangent being drawn, it shall be equal to the diameter. Let the distance sought be x, the diameter = d, and the tangent also = d, then (d+ x)x = d2, or x2 + dx = ď2; 19 .. x= (√5-1); or the greater segment of the diameter divided medially. |