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. = B CI B' с α If we take SA' = SA, and through the points A and A' the A parallel planes ABCD and A'B'C' D' be passed, we shall have S B' = SB, SC' SC, etc. For if we conceive a third parallel plane to pass through S, then A A', BB', etc., are divided proportionally, § 469. And if any one bisected at S, the others are also bisected at S. points A', B', etc., are symmetrical with A, B, etc. of them be Hence, the Moreover, the two symmetrical polyhedral angles are equal in all their parts. For their face angles ASB and A' SB', 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 abcd, in the same plane with ABCD, it will be evident that while the parts AS B, 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. A D B Let S-ABC be a trihedral angle in which the face angle ASC is greater than either angle ASB or angle BSC. LASB+Z BSC > < ASC. = We are to prove cutting AS and SC. :. ZA SB + Z BSC > ≤ ASD+Z DSC, that is ZASB+L B SC > ZASC. Iden. Cons. Cons. § 106 Cons. Iden. § 116 Q. E. D. 488. The sum of the face angles of any convex polyhedral angle is less than four right angles. S E D B Let the polyhedral angle S be cut by a plane, making the section ABC DE a convex polygon. We are to prove ASB+ 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 O will be the same as the number having their common vertex at S. .. the sum of all the 4 of the ▲ having the common vertex at S is equal to the sum of all the of the ▲ having the common vertex at 0. But in the trihedral formed at A, B, C, etc. ZSAE+ZSAB> ZOAE+ZOAB, § 487 (the sum of any two face of a trihedral is greater than the third). SBC > ZOBA+Z OBC. § 487 and 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 O. PROPOSITION XXV. THEOREM. 489. An isosceles trihedral angle and its symmetrical trihedral angle are equal. B' Let S-ABC be an isosceles trihedral angle, having LASB LBSC. And let S-A'B'C' be its sym metrical trihedral angle. We are to prove trihedral ZS-ABC=trihedral ▲ S-A'B'C'. Revolve S-A'B'C' about S until SB' falls on SB and the plane SB'A' falls on the plane SB C. Now the dihedral / C-S B-A = dihedral A'-S B'-C', (being vertical dihedral ▲). Now and .. the plane SB'C' will fall on the plane SB A. In like manner SC will fall on SA, .. the two trihedral s will coincide and be equal. Q. E. D. PROPOSITION XXVI. THEOREM. 490. Two symmetrical trihedral angles are equivalent. Let the trihedral S-ABC and S-A'B'C' be symmetrical. We are to prove trihedral ZS-ABC* trihedral ZS-A'B'C'. Draw D'D making the DSA, DSC, and DSB equal. Then = (being vertical of the equal & DSA, DSC, and D SB). Then the trihedral ≤ S-DCB=trihedral ≤ S-D' C' B', (two isosceles symmetrical trihedral ▲ are equal). $489 Adding the first two equalities, the polyhedral ZS-ABCD = polyhedral S-A'B'C' D'. Take away from each of these equals the equal trihedral S-ADB and S-A' D' B'. Then trihedral Z S-A BC trihedral ▲ S-A' B' C'. Q. E. D. 491. SCHOLIUM. If DD' fall within the given trihedral angles these trihedral angles would be composed of three isosceles trihedral angles which would be respectively equal, and hence the given trihedral angles would be equivalent. * The symbol (~) is to be read "equivalent to." |