PROPOSITION XXII. THEOREM. 479. If two straight lines be not in the same plane, one and only one common perpendicular to the lines can be drawn. Let A B and C D be two given straight lines not in the same plane. We are to prove one and only one common perpendicular to the two lines can be drawn. not ll, Since AB and CD are not in the same plane they are (two Ils lie in the same plane). $ 474 Through the line A B pass the plane MN to CD. Since CD is to the plane MN, all its points are equally distant from the plane M N ; hence C'D', the projection of the line CD on the plane M N, will be | to CD, § 432 § 76 and will intersect the line A B at some point as C'. Now since CC' is the line which projects the point Cupon the plane MN, it is to the plane MN; hence CC is (if a line be to a plane, it is be to C'D' and A B, $435 $430 Also, CC' is to CD, .. CC is the common to the lines CD and A B. Moreover, line C C is the only common 1. § 67 For, if another line E B, drawn between A B and C D, could to AB and C D, it would also be to a line BG drawn Il to CD in the plane M N, and hence But EH, drawn in the to the plane M N. $ 67 § 449 plane MN. plane C D' to C C', is to the Hence we should have two § 457 MN, which is impossible, .. CC' is the only common to the lines CD and A B. from the point E to the plane $447 Q. E. D. ON POLYHEDRAL ANGLES. 480. DEF. A Polyhedral angle is the extent of opening of three or more planes meeting in a common point. B E The plane angles ASB, BSC, etc., formed by the edges are its face angles. 481. DEF. Polyhedral angles are classified as trihedral, quadrahedral, etc., according to the number of the faces. 482. DEF. Trihedral angles are rectangular, bi-rectangular, or tri-rectangular, according as they have one, two, or three right dihedral angles. 483. DEF. Trihedral angles are scalene, isosceles, or equilateral, according as the face angles are all unequal, two equal, or three equal. 484. DEF. A polyhedral angle is convex, if the polygon formed by the intersections of a plane with all its faces be a convex polygon. 485. DEF. Two polyhedral angles are equal when they can be applied to each other so as to coincide in all their parts. Since two equal polyhedral angles coincide however far their edges and faces be produced, the magnitude of a polyhedral angle does not depend upon the extent of its faces. But, in order to represent the angle in a diagram, it is usual to pass a plane, as ABCDE, cutting all its faces in the straight lines, A B, BC, etc.; and by the face ASB is meant the indefinite surface included between the lines SA and SB indefinitely produced. 486. DEF. Two polyhedral angles are symmetrical if they have the same number of faces, and the successive dihedral and face angles respectively equal but arranged in reverse order. Thus, if the edges AS, BS, etc., of the polyhedral angle, S-ABCD, be produced, there is formed another polyhedral angle, S-A'B'C' D', which is symmetrical with the first, the vertex S being the centre of symmetry. If we take SA' SA, and through the points 4 and 4' the A parallel planes ABCD and A'B'C' D' be passed, we shall have SB' = = SB, SC SC, etc. B B C For if we conceive a third Hence, the parallel plane to pass through S, then A A', B B', etc., are divided proportionally, § 469. And if any one of them be bisected at S, the others are also bisected at S. points A', B', etc., are symmetrical with A, B, etc. Moreover, the two symmetrical polyhedral angles are equal in all their parts. For their face angles ASB and A'S B', BSC and B'SC are equal each to each, being vertical plane angles. And the dihedral angles formed at the edges SA and SA', SB and SB', are equal each to each, being vertical dihedral angles. Now if the polyhedral angle S-A'B'C' D' be revolved about the vertex S until the polygon A'B'C' D' is brought into the position a bed, in the same plane with ABCD, it will be evident that while the parts ASB, B SC, etc., succeed each other in the order from left to right, the corresponding equal parts a Sb, b Sc, etc., succeed each other in the order from right to left. Hence the two figures cannot be made to coincide by superposition, but are said to be equal by symmetry. PROPOSITION XXIII. THEOREM. 487. The sum of any two face angles of a trihedral angle is greater than the third. S Let S-ABC be a trihedral angle in which the face angle ASC is greater than either angle ASB or angle BSC. We are to prove Z ASB+Z BSC > ZASC. = In the face A SC draw SD, making ▲ AS D ZAS B. Pass a plane through A C and the point B. (being homologous sides of equal ▲). In the AA BC, AB+BC > A C. Then Subtract the equals A B and A D. Now in the ABSC and DSC :. ZA SB + Z BSC > ZA SD + 2 DSC, that is ZASB+ZBSC>LASC. Cons. Iden. § 116 QE D 488. The sum of the face angles of any convex polyhedral angle is less than four right angles. Let the polyhedral angle S be cut by a plane, making the section A B C D E a convex polygon. We are to prove ASB+Z BSC etc. < 4 rt. ¿. From any point O within the polygon draw O A, O B, O C, OD, O E. The number of the A having their common vertex at 0 will be the same as the number having their common vertex at S. .. the sum of all the of the ▲ having the common vertex at S is equal to the sum of all the of the A having the common vertex at 0. and But in the trihedral formed at A, B, C, etc. ZSAE+Z SAB>ZOAE+ZOA B, $ 487 § 487 (the sum of any two face of a trihedral ▲ is greater than the third). SBC > ZOBA+Z OBC. SBA + .. the sum of the at the bases of the A whose common vertex is S is greater than the sum of the at the bases of the A whose common vertex is 0. .. the sum of the at S is less than the sum of the at 0. |