BOOK VII. PLANES. ·DIEDRAL AND POLYEDRAL ANGLES. DEFINITIONS. 388. A STRAIGHT line is perpendicular to a plane, when it is perpendicular to every straight line which it meets in that plane. Conversely, the plane, in the same case, is perpendicular to the line. Α M B N The foot of the perpendicular is the point in which it meets the plane. Thus the straight line A B is perpendicular to the plane MN; the plane MN is perpendicular to the straight line AB; and B is the foot of the perpendicular A B. 389. A line is parallel to a plane when it cannot meet the plane, however far both of them may be produced. Conversely, the plane, in the same case, is parallel to the line. M 390. Two planes are parallel to each other, when they cannot meet, however far both of them may be produced. 391. A DIEDRAL ANGLE is an angle formed by the intersection of two planes, and is measured by the inclination of two straight lines drawn from any point in the line of intersection, perpendicular to that line, one being drawn in each plane. A B Z The line of common section is called the edge, and the two planes are called the faces, of the diedral angle. Thus the two planes A B M, A ABN, whose line of intersection is A B, form a diedral M N angle, of which the line AB B is the edge, and the planes ABM, ABN are the faces. 392. A diedral angle may be acute, right, or obtuse. If the two faces are perpendicular to each other, the angle is right. 393. A POLYEDRAL ANGLE is an angle formed by the meeting, at one point of more than two plane angles, which are not in the same plane. The common point of meeting of the planes is called the vertex, each of the plane angles a face, A B S C and the line of common section of any two of the planes an edge of the polyedral angle. Thus the three plane angles ASB, BS C, CSA form a polyedral angle, whose vertex is S, whose faces are the plane angles, and whose edges are the sides, AS, BS, CS, of the same angles. 394. A polyedral angle formed by three faces is called a triedral angle; by four faces, a tetraedral; by five faces, a pentaedral, &c. PROPOSITION I.-THEOREM. '395. A straight line cannot be partly in a plane, and partly out of it. For, by the definition of a plane (Art. 10), a straight line which has two points in common with a plane lies. wholly in that plane. 396. Scholium. To determine whether a surface is a plane, apply a straight line in different directions to that surface, and ascertain whether the line throughout its whole extent touches the surface. PROPOSITION II. - THEOREM. 397. Two straight lines which intersect each other lie in the same plane and determine its position. Let AB, AC be two straight lines which intersect each other in A; then these lines will be in the same plane. B A Conceive a plane to pass through A B, and to be turned about AB, until it pass through the point C; then, the two points A and C being in this plane, the line AC lies wholly in it (Art. 10). Hence, the position of the plane is determined by the condition of its containing the two straight lines A B, A C. 398. Cor. 1. A triangle, A B C, or three points, A, B, C, not in a straight line, determine the position of a plane. 399. Cor. 2. Hence, also, two parallels, A B, 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. A C E F B Ꭰ 400. If two planes cut each other, their common section is a straight line. Let the two planes A B, CD cut each other, and let E, F be two points in their common section. Draw the straight line EF. Now, since the points E and F are in the plane A B, and also in the plane CD, the straight line EF, joining E and F, must be wholly in each plane, or is common to both of them. Therefore, the common section of the two planes AB, A CD is a straight line. PROPOSITION IV.-THEOREM. 401. If a straight line is perpendicular to each of two straight lines, at their point of intersection, it is perpendicular to the plane in which the two lines lie. Let the straight line AB be perpendicular to each of the straight lines CD, EF, at B, the point of their intersection, and MN the plane in which the lines CD, EF lie; then will A B be perpendicular to the plane MN. M A F B E D Through the point B draw any straight line, BG, in the plane MN; and through any point G draw DG F, meeting the lines CD, EF in such a manner that DG shall be equal to GF (Prob. XXVIII. Bk. V.). Join AD, AG, AF. The line D F being divided into two equal parts at the point G, the triangle D B F gives (Prop. XIV. Bk. IV.) B F2 + B D2 = 2 B G2 + 2 G F2. + Ꭰ The triangle DAF, in like manner, gives A F2 + A D2 = 2 AG2 + 2 G F2. Subtracting the first equation from the second, and ob serving that the triangles A B F, ABD, each being rightangled at B, give Ꭰ A F2 — B F2 — AB, and AD - B D2 — A B2, we shall have = A B+ AB = 2 AG2 - 2 BG2. = Therefore, by taking the halves of both members, we have A B2 = A G2 — BG3, or 2 A G2 = A B2+BG2; Ᏼ hence, the triangle A B G is right-angled at B, and the side A B is perpendicular to BG. In the same manner, it may be shown that A B is perpendicular to any other straight line in the plane MN, which it may meet at B; therefore AB is perpendicular to the plane MN (Art. 388). 402. Scholium. Thus it is evident, not only that a straight line may be perpendicular to all the straight lines. which pass through its foot, in a plane, but it always must be so whenever it is perpendicular to two straight lines drawn in the plane; which shows the accuracy of the first definition (Art. 388). 403. Cor. 1. The perpendicular AB is shorter than any oblique line A G; therefore it measures the shortest distance from the point A to the plane M N. 404. Cor. 2. From any given point, B, in a plane, only one perpendicular to that plane can be drawn. For if there could be two, conceive a plane to pass through them, intersecting the plane MN in BG; the two perpendiculars would then be perpendicular to the straight line BG at the same point, and in the same plane, which is impossible (Prop. XIII. Cor., Bk. I.). It is also impossible to let fall from a given point out of a plane two perpendiculars to that plane. For, suppose A B, A G to be two such perpendiculars, then the triangle ABG will have two right angles, A B G, A GB, which is impossible (Prop. XXVIII. Cor. 3, Bk. I.). |