PROBLEM XIII. Having given two polygons, P and Q, to construct a third polygon X similar to the first, and such that the second may be to the third in the ratio of m to n. Let a denote a side of the polygon P, and a the homologous side of the polygon X. Then we shall have, by the conditions of the problem, P:X::α2: x2, This being supposed, conceive the polygons P and Q transformed into equivalent squares (P. VIII.), having respectively for their sides p and q. In the same manner let y represent the side of a square equivalent to X; the preceding proportions will be changed into the following: Now, the last of these proportions will make known y by the construction of Problem XI., and the second will then give x as a fourth proportional to the lines p, y, and a, which will be the side of the polygon sought, homologous to a. PROBLEM XIV. To construct a rectangle equivalent to a given square m2, and such that the sum or the difference of two adjacent sides may be equal to a given line a. First case. The given line a being the sum of the adjacent sides. C K A Upon a straight line AB = a, as a diameter, describe a circumference; through A the extremity of the diameter AB, draw the perpendicular AC=m. Through ▲ C draw CL parallel to AB, meeting the circumference in the points E, E'. From the points E and E' draw EF, E'F' perpendicular to AB. The distances AF and FB, or AF' and F'B, will be the two adjacent sides of the rectangle required to be constructed. In effect, we have (T. I., S. I.), and AF × FB=EF2 = AC2 = m2, AF+FBAB= a. The problem is impossible if AC> OI or OA; that is, when m > 1a. If m = a, the points E, E' will unite at I, and the rectangle will become a square. This demonstrates that the greatest rectangle which we can form by dividing a given line into two parts for the adjacent sides, is the square constructed on half the given line. Second case. The given line being the difference of two adjacent sides. After having drawn AC=m, as in the first case, we join C and O, and produce it so as to meet the circumference at the points K and G. The two lines CG, CK will be the adjacent sides of the rectangle sought. We have, by the construction (T. III.), and 2 CA2= CG × CK, or CG × CK= m2, CG-CKKG=AB= a. In this case, the problem is always possible for all values of a and m. FIFTH BOOK. OF THE PLANE, AND ITS COMBINATION WITH THE STRAIGHT LINE. DEFINITIONS. I. The intersection of two planes is the line in which they meet to cut each other. It is obvious, from our definition of a plane (B. I., D. XVII.), that this intersection is a straight line. II. A line is perpendicular to a plane, when it is perpendicular to all the lines in that plane which meet it. III. One plane is perpendicular to another, when every line in the one which is perpendicular to their intersection is perpendicular to the other plane. IV. A line is parallel to a plane, when, if both are produced to any distance, they do not meet; and, conversely, the plane is then also parallel to the line. V. Two planes are parallel to each other, when, both being produced to any distance, they do not meet. VI. The inclination of two planes which intersect, is called a diedral angle. The planes forming a diedral angle are called the faces of the angle, and their intersection is called its edge. If a rectangle ABCD is revolved about its side AB as an axis, so as to assume in succession the positions ABEF, ABGH, etc., the point C will describe the circumference of a circle. In fact, any point of the plane of this rectangle, not situated in the axis AB, will, in a similar manner, describe the circumfer B A H F E G D ence of a circle having its centre in the axis. During this revolution, the arc CE will increase in the same ratio as the opening or inclination of the planes ABCD, ABEF, increases; that is, as the diedral angle DABE increases. Now, the arc CE is the measure of the angle CBE (B. II., T. IX., S.), hence a diedral angle has for its measure the angle contained by two lines drawn from any point of its edge, perpendicular to the same, one in each face. VII. When several planes pass through a common point, the angular space included between these planes is called a polyedral angle. Each plane is called a face; the line in which any two faces intersect is called an edge; and the common point through which the planes pass is called the vertex. Three planes, at least, are necessary for forming a polyedral angle. In the case of three planes, the angle is called a triedral angle. THEOREM I. One part of a straight line cannot be in a plane, and another part out of it. For (B. I., D. XVII.), when a straight line has two points common with a plane, it must coincide with the plane. THEOREM II. Two straight lines which intersect each other, are situated in the same plane, and determine its position. plane, then it is clear that the two lines, according to the terms of the proposition, are in the same plane; but if not, let the plane passing through AB be supposed to be turned round AB till it passes through the point C, then the line AC, which has two of its points, A and C, in this plane, coincides with it; and hence the position of the plane is determined by the single condition of containing the two straight lines AB, AC. Cor. I. A triangle ABC, or any three points not in a straight line, determines the position of a plane. Cor. II. Hence, also, two parallels AB, CD determine the position of a plane; for, drawing the secant EF, the plane of the two straight lines AB, EF is that of the parallels AB, CD. THEOREM III. The intersection of two planes is a straight line. Let DC and EF be two planes cutting each other, and A, B two points in which the planes meet. Draw the straight line AB; this line is the intersection of the two planes. For, since the straight line touches the two planes in the points A and B, it lies wholly in both these planes, or is common to both of them; that is, the intersection of the two planes is in a straight line. D If THEOREM IV. a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of these lines. Let AP be perpendicular to the two lines PB, PC, at their point of intersection P; then will it be perpendicular to MN the plane of the lines. N A M B P C Through P draw in the plane MN any line, as PQ; and through any point of this line, as Q, draw BQC, so that BQ = QC (B. IV., P. III., C. II.); join AB, AQ, AC. The base BC being divided into two equal parts at the point Q, the triangle BPC (B. III., T. XVII.) will give PC2+PB2 = 2PQ2+2QC2. |