can be drawn from A, falling on the same side of BA with CD, and not cutting the circle, which shall not cut CD. This second tangent, then, must be either AD or some other line, AM, cutting CD in a point M differing from D. If now AM is a tangent, ABCM is a circumscribed quadrilateral, and by the preceding proposition we shall have But we also have, by the hypothesis of the present proposition, AB+DC=AD+BC. Taking the difference of these equalities, we have M D с that is, one side of a triangle is equal to the difference of the other two, which is absurd. Therefore, the hypothesis that the tangent drawn from A and cutting the line CD, cuts it in any other point than D, leads to an absurdity; therefore, that hypothesis must be false, and the tangent in question must cut CD in D, and consequently coincide with AD. Hence, a circle has been described which is tangent to the four sides of the quadrilateral; and the quadrilateral is circum scribed about the circle. 103. Scholium. The method of demonstration employed above is called the indirect method, or the reductio ad absurdum. At the outset of a demonstration, or at any stage of its progress, two or more hypotheses respecting the quantities under consideration may be admissible so far as has been proved up to that point. If, now, these hypotheses are such that one must be true, and only one can be true, then, when all except one are shown to be absurd, that one must stand as the truth. While admitting the validity of this method, geometers usually prefer the direct method whenever it is applicable. There are, however, propositions, such as the preceding, of which no direct proof is known, or at least no proof sufficiently simple to be admitted into elementary geometry. We have already employed the reductio ad absurdum in several cases without presenting the argument in full; see (I. 47), (I. 85), (27). Dian BOOK III. PROPORTIONAL LINES. SIMILAR FIGURES. THEORY OF PROPORTION. 1. DEFINITION. One quantity is said to be proportional to another when the ratio of any two values, A and B, of the first, is equal to the ratio of the two corresponding values, A' and B', of the second; so that the four values form the proportion This definition presupposes two quantities, each of which can have various values, so related to each other that each value of one corresponds to a value of the other. An example occurs in the case of an angle at the centre of a circle and its intercepted arc. The angle may vary, and with it also the arc; but to each value of the angle there corresponds a certain value of the arc. It has been proved (II. 51) that the ratio of any two values of the angle is equal to the ratio of the two corresponding values of the arc; and in accordance with the definition just given, this proposition would be briefly expressed as follows: "The angle at the centre of a circle is proportional to its intercepted arc." 2. Definition. One quantity is said to be reciprocally proportional to another when the ratio of two values, A and B, of the first, is equal to the reciprocal of the ratio of the two corresponding values, A' and B', of the second, so that the four values form the proportion PROPOSITION XXXVIII.-PROBLEM. 89. At a given point in a given circumference, to draw a tangent to the circumference. Let A be the given point in the given circumference. Draw the radius OA, and at A draw BAC perpendicular to OA; BC will be the required tangent (26). If the centre of the circumference is not given, it may first be found by the preceding problem, or we may proceed more directly as follows. Take two points D and E equidistant from A; draw the chord DE, and through A draw BAC parallel to DE. Since A is the middle point of the arc DE, the radius drawn to A will be perpendicular to DE (16), and consequently also to BC; therefore BC is a tangent at A. B 0 B E PROPOSITION XXXIX.-PROBLEM. 90. Through a given point without a given circle to draw a tangent to the circle.. Let O be the centre of the given circle and P the given point. Upon OP, as a diameter, describe a circumference intersecting the circumference of the given circle in two points, A and A'. Draw PA and PA', both of which will be tangent to the given circle For drawing the radii OA and OA', the angles OAP and OA'P are right angles (59); therefore PA and PA' are tangents (26). In practice, this problem is accurately solved by placing the straight edge of a ruler through the given point and tangent to the given circumference, and then tracing the tangent by the straight edge. The precise point of tangency is then determined by drawing a perpendicular to the tangent from the centre. 91. Scholium. This problem always admits of two solutions. Moreover, the portions of the two tangents intercepted between the given point and the points of tangency are equal, for the right triangles POA and POA' are equal (I. 83); therefore, PA = PA'. PROPOSITION XL.-PROBLEM. 92. To draw a common tangent to two given circles. Let O and O' be the centres of the given circles, and let the radius of the first be the greater. 1st. To draw an exterior common tangent. With the centre O, and a radius OM, equal to the difference of the given radii, describe a circumference; and from O' draw a tangent O'M to this circumference (90). Join OM, and produce it to meet the given circumference in A. Draw O'A' parallel to = B' B OA, and join AA'. Then AA' is a common tangent to the two given circles. For, by the construction, OMOA — O'A', and also OMOA ·MA, whence MA O'A', and AMO'A' is a parallelogram (I. 108). But the angle M is a right angle; therefore, this parallelogram is a rectangle, and the angles at A and A' are right angles. Hence, AA' is a tangent to both circles. Since two tangents can be drawn from O' to the circle OM, there are two exterior common tangents to the given circles, namely, AA' and BB', which meet in a point T in the line of centres 00' produced. 2d. To draw an interior common tangent. With the centre 0 and a radius OM equal to the sum of the given radii, describe a circumference, and from O' draw a tangent O'M to this circumference. Join OM, intersecting the given circumference in A. Draw O'A' parallel to OA. Then, since OM = OA + O'A', we have AM O'A', and AMO'A' is a rectangle. Therefore, AA' is a tangent to both the given circles. There are two interior common M T parts, and let A contain m of these parts with a remainder less than one of the parts; then we have 1 and, since may be taken as great as we please, may be made less m n than any assigned measure of precision, and will be the approxi n within that assigned measure. 49. Theorem. Two incommensurable ratios are equal, if their approxi mate numerical values are always equal, when both are expressed within the same measure of precision however small. A Let and be two incommensurable ratios whose approximate B A' numerical values are always the same when the same measure of precision is employed in expressing both; then, we say that 1 n For, let be any assumed measure of precision, and in accordance with the hypothesis of the theorem, suppose that for any value of 1 n then, these ratios cannot differ from each other by so much as 1 But the measure may be assumed as small as we please, that is less n than any assignable quantity however small; hence and cannot differ by any assignable quantity however small, and therefore they must be equal. The student should study this demonstration in connection with that of Proposition XIX., which follows. |