the angles EMA, EM'A being in a semicircle, are right (T., X. C. I.). Consequently, AM and AM' are respectively perpendicular to the radii EM and EM', and therefore tangent to the circumference (T. V.). Scholium. The case when the given point is in the circumference offers no difficulty, since we then draw a radius to the given point, and through its extremity draw a perpendicular. OF COMMON MEASURE. PROBLEM XIII. To find the common measure of two given lines, provided they have one, and consequently their numerical ratio. Let AB and CD be the given lines. From the greater AB cut off parts equal to the less line CD, as many times as possible; for example, twice, with the remainder FB. From the line CD cut off parts equal to FB, as many times as possible; for example, once, with the remainder GD. From the first remainder FB cut off parts equal to the remainder GD, as many times as possible; for example, once, with the remainder HB. A G D F H B From the second remainder GD cut off parts equal to the third remainder HB, as many times as possible; for example, twice, without a remainder. The last remainder, HB, will be a common measure of the given lines. If we regard HB as a unit, GD will be 2, and FB=FH+HB=GD+HB=3; CD=CG + GD = FB+GD=3+2=5; AB= AF+FB=2 CD + FB=10+3=13. Therefore the line AB is to the line CD as 13 to 5. If AB is taken for the unit, CD will be; but if CD be taken as the unit, AB will be 13 5 13 5 26 If AB is of a yard, then CD will be of a yard, or of a yard. 13 Again, if CD is of a foot, then AB will be 13 of 3 of a foot = 28 of a foot; and so on for other comparisons. Cor. Since arcs of the same circle, or of equal circles, are capable of superposition, this same method may be employed when we wish to find a common measure of two arcs of the same circle, or of equal circles. Scholium. Magnitudes frequently have no common measure; that is, they are incommensurable; in which case, we shall always, by the foregoing method, have a remainder, however far we carry our successive operations. But these successive remainders would finally become so small that they might practically be neglected. Consequently, we can always find two numbers whose ratio shall approximate as close as we please to the ratio of the given incommensurable magnitudes. This being admitted, we say that two incommensurable magnitudes A and B are proportional to two other incommensurable magnitudes A' and B', or that the ratio of A to B is equal to the ratio of A' to B', when we obtain for each the same numerical ratio at each successive step of the approximation. PROBLEM XIV. Show, geometrically, that the side of a square and its diagonal are incommensurable. Let ABCD be the square, and AC its diagonal. Cutting off AF from the diagonal equal to AB, a side of the square, we have the remainder CF, which must be compared with CB. If we join FB, and draw FG perpendicular to AC, the triangle BGF will be isosceles. D A F C B For, the angle ABG = AFG, each being a right angle; and since the triangle ABF is isosceles, the angle ABF=AFB. Therefore, subtracting the angle ABF from ABG, the remainder FBG will equal the angle BFG, found by subtracting the angle AFB from AFG. Consequently the triangle BGF is isosceles, and BGFG; but, since AC is the diagonal of a square, the angle FCG is half a right angle; but CFG is a right angle, and consequently FGC is also half a right angle, and CG is the diagonal of a square whose side is CF. Hence, after CF = FG = BG has been taken once from CB, it remains to take CF from CG; that is, to compare the side of a square with its diagonal, which is the very question we set out with, and of course we shall find precisely the same difficulty in the next step of the process; so that, continue as far as we please, we shall never arrive at a term in which there will be no remainder. Therefore there is no common measure of the diagonal and side of a square. THIRD BOOK. THE PROPORTIONS OF STRAIGHT LINES AND THE AREAS OF RECTILINEAL FIGURES. DEFINITIONS. I. Two figures are similar when they resemble each other in all their parts, differing only in their comparative magnitudes, so that every point of the one has a corresponding point in the other. If, in two similar figures, we conceive a line joining any two points of the one, and another line joining the corresponding points of the other, the numerical ratio of these lines, which is always the same, is called the ratio of similitude. When the ratio of similitude is equal to unity, the two similar figures become equal in all respects. II. Two triangles are similar when their corresponding sides are proportional. III. Two polygons are similar when they are capable of being decomposed into the same number of similar triangles, each to each, having the same order of arrangement. IV. In similar figures homologous points, lines, or angles, are those which have like positions, in regard to the two figures. V. The area of a surface, or its superficial extent, is the nu merical ratio of this surface referred to its unit surface. We readily see that figures of very different forms may have the same superficial extent. When we wish to express the property that two surfaces have the same area, without, however, being equal, or capable of superposition, we say they are equivalent. VI. In determining the area of surfaces, it is found convenient to call the bases of a parallelogram the sides which are parallel. The altitude or height is the common perpendicular to these bases, of which one is called the inferior or lower base, and the other the superior or upper base. In the rectangle, two consecutive sides form the base and the height respectively. In the square, the base and height are equal. In the case of a triangle, either side may be regarded as the base. The height will be the perpendicular, drawn from the opposite angle to the base, or to the base produced. PROPORTIONAL LINES. THEOREM I. If equal portions are taken on a straight line, and through the points of division parallel lines are drawn meeting a second line, the intercepts of this second line will be equal also. If the portions CD, DE, and EF, of the straight line AB, are equal; and if through the points C, D, E, and F, the parallels CC', DD', EE', and FF', are drawn, meeting the line A'B', then will the intercepts C'D', D'E', and EF' be equal. For, drawing Cc, Dd, and Ee, parallel to A'B', the triangles CDc, F D с A A' E d B C' D' E/ F/ B' DEd, and EFe will be equal, since the sides CD, DE, and EF are equal, and their corresponding angles are obviously equal (B. I., T. III.). Hence, Cc Dd= Ee, and consequently C'D'= = D'E'= E'F' (B. I., T. XXXI., C. II.). THEOREM II. If a line is drawn parallel to the bases of a trapezoid, intersecting the sides, it will divide them into proportional parts. If EF is parallel to AD and BC, we shall have AE: EB:: DF: FC. First. When AE and EB are commensurable; that is, when they are to B E |