Proposition 240 Theorem Between two skew lines one common perpendicular can be drawn, and only one. Hypothesis. AB and CD are two skew lines. Conclusion. One common to AB and CD can be drawn, and only one. Proof. Through CD pass the plane MN | to AB. Through AB pass the plane AE to MN, intersecting MN in the line FE. FE is to AB. FE is not to CD. Let F be the intersection of CD and FE. At F erect FA 1 to FE in the plane AE. (?) (?) (?) (?) (?) Hence FA is a common 1 to AB and CD. If possible, suppose that GH is another common to AB and CD. Let the plane determined by AB and GH intersect MN in the line HK. But GR, drawn in the plane AE 1 to FE, is to MN. (?) Then there are two is from G to MN. But this is impossible. (?) .. GH is not a common to AB and CD, and FA is the only common 1. Q.E.D. COR. The common perpendicular is the shortest line that can be drawn between two skew lines. GR < GH; ..AF < GH. POLYHEDRAL ANGLES When three or more planes meet at a common point, they are said to form a polyhedral angle, or solid angle. The common point is called the vertex of the angle, the intersections of the planes are called the edges of the angles, the portions of the planes included between the edges are called the faces, and the angles formed by the edges are called the face angles. For example, the planes BAC, CAD, DAE, and EAB, meeting at the common point A, form the polyhedral angle A-BCDE. A is the vertex; AB, AC, AD, and AE are the edges; the planes BAC, CAD, DAE, and EAB are the faces; and the angles BAC, CAD, DAE, and EAB are the face angles. A polyhedral angle of.three faces is called B a trihedral angle; a polyhedral angle of four faces is called a tetrahedral angle, and so on. A trihedral angle is called rectangular, bi-rectangular, or tri-rectangular, according as it has one, two, or three right dihedral angles. A trihedral angle is called isosceles when two of its face angles are equal. Two polyhedral angles are called vertical when they have a common vertex and the edges of one are the prolongations of the edges of the other. The polygon formed by the intersections of the faces of a polyhedral angle with a plane cutting all the edges is called a section of the polyhedral angle. In the figure on page 307, BCDE is a section of the polyhedral angle A-BCDE. A polyhedral angle is said to be convex when any section is a convex polygon. The face angles and the dihedral angles formed by the faces are called the parts of a polyhedral angle. Two polyhedral angles are congruent when the parts of one are equal respectively to the parts of the other and are arranged in the same order, for one may be applied to the other so that they will coincide. Two polyhedral angles are said to be symmetrical when the parts of one are equal respectively to the parts of the other, but are arranged in reverse order. In general, two symmetrical polyhedral angles do not coincide when one is applied to the other. Many properties of trihedral angles are analogous to properties of triangles. A number of theorems of Plane Geometry concerning triangles may be changed to theorems concerning trihedral angles by changing angle and side to dihedral angle and face angle respectively. Two vertical polyhedral angles are symmetrical. Consult Prop. 5 and Prop. 229, Cor. II. Ex. 1019. In an isosceles trihedral angle, the dihedral angles opposite the equal face angles are equal. Ex. 1020. State and prove the converse of Ex. 1019. Proposition 242 Theorem Two trihedral angles are either congruent or symmetrical when two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other. CASE I. When the equal parts are arranged in the same order, the trihedral angles are congruent. A A' B Hypothesis. In the trihedral B' A-BCD and A'-B'C'D', ▲ BAD = LB'A'D', ▲ BAC = B'A'C', and dihedral LAB = dihedral A'B'. Conclusion. Trihedral A-BCD = trihedral A'-B'C'D'. Proof. Apply A-BCD to A'-B'C'D' so that ▲ BAD shall coincide with B'A'D', AB falling along A'B' and AD along A'D'. Face BAC will fall on face B'A'C'. Face CAD will fall on face C'A'D'. (?) (?) (?) .. trihedral A-BCD = trihedral A'-B'C'D'. (?) CASE II. When the equal parts are arranged in reverse order, the trihedral angles are symmetrical. HINT. Produce the edges of one of the given trihedral angles through the vertex, thus forming a trihedral angle which can be proved congruent to the other given trihedral angle. Two trihedral angles are either congruent or symmetrical when a face angle and the two adjacent dihedral angles of one are equal respectively to a face angle and the two adjacent dihedral angles of the other. If two trihedral angles have the three face angles of one equal respectively to the three face angles of the other, the homologous dihedral angles are equal. A A B Hypothesis. In the trihedral 4A-BCD and A'-B'C'D', ▲ BAC = B'A'C', ▲ CAD = C'A'D', and ▲ BAD = ▲ LB'A'D', Conclusion. = Dihedral AB = dihedral ZAC dihedral A'D' A'C', and dihedral Proof. On the edges take AB Draw BC, CD, BD, B'C', C'D', and B'D'. A BACA B'A'C' .. BC = B'C' AD AD In like manner it can be proved that BD CD = C'D'. .. A BCD = ▲ B'C'D'. = = B'D' and (?) |