CHAMBERS'S INFORMATION FOR THE PEOPLE. 



angle ABCD is measured by the plane angle 

 POQ, where O is any point in BC, and PO, OQ, 

 lines drawn through this point perpendicular to 

 BC, and in the planes AC and DC. 



PROP. XV. Every plane passing through a 

 perpendicular to a plane, is also perpendicular to 

 the plane. 



Let the line AB be perpendicular to the plane 

 P, then any plane CD which 

 passes through AB is also 

 perpendicular to P. From 

 the point B draw in the 

 plane P, BE perpendicular 

 to BD ; then since AB is 

 perpendicular to BD, the 

 dihedral angle of the two 

 planes is measured by the plane angle ABE. 

 But ABE is a right angle, since AB is by hypoth- 

 esis perpendicular to the plane P. Hence the 

 plane CD is perpendicular 

 to the given plane P. 



Cor. It follows that if 

 two planes intersecting be 

 each perpendicular to a 

 third plane, the line of their 

 intersection is also perpen- 

 dicular to the third plane. 



PROP. XVI. To two straight lines in space not 

 being in the same plane one common perpendicular 

 can be drawn, and this is the shortest line joining 

 the given lines. 



Let AB and CD be the two straight lines not in 

 the same plane. Through A draw AE parallel to 

 CD, and drop a perpendicular DF to the plane of 

 AEB, and let the plane of 

 CDF cut the plane AEB 

 in the line FG intersect- 

 ing AB in G. Draw GH 

 parallel to FD, to meet A 

 CD in H ; GH is the 

 line required. For since 

 AE and CD are parallel, 

 CD is parallel to the plane AEB (Prop. IX., p. 

 628), and therefore GF is parallel to CD ; there- 

 fore DF is perpendicular to CD. Therefore, also, 

 GH is perpendicular to CD and to the plane 

 AEB, and GH is perpendicular both to CD and 

 AB. Also, if any other line DK be drawn to join 

 the given lines, it is greater than GH ; for DK -7 

 DF, and DF = HG, therefore DK T HG. 



TRIHEDRAL ANGLES. 



Definition. The figure formed by several planes 

 meeting in one point is called 

 a polyhedral angle. 



Thus the planes SAB, SBC, 

 SCO, SDE, SEA, which pass 

 through the same point S, and 

 are terminated by their inter- 

 sections SA, SB, SC, SD, 

 SE, form a polyhedral angle, 

 and the point S is the vertex, 

 and SA, SB, SC, &c., the 

 edges. The. plane angles 

 ASB, BSC, CSD, &c., are 

 called the faces of the polyhedral S. 



630 



Three planes at least are necessary to form a 

 polyhedral angle, and a poly- 

 hedral angle which has only three 

 faces is termed a trihedral angle. 

 Thus the figure SABC is called a 

 trihedral angle. 



A polyhedral angle is convex 

 or concave, according as it lies 

 on the same or on different sides 

 of each of the planes produced 

 which form it 



Two polyhedral angles are said 

 to be vertically opposite when 

 the edges of the one are the con- 

 tinuations of the edges of the 

 other; thus the two polyhedral 

 angles SABCDE, SA'B'C'D'E' 

 are vertically opposite. And it 

 will be at once evident that, 

 although the dihedral angles and 

 also the plane angles of these 

 figures are equal, the figures are 

 not superposable. Such angles are 

 said to be symmetrical. 



PROP. XVII. In any trihedral angle, tJif 

 sum of any two of its faces is greater than Hie 

 third. 



Let SABC be a trihedral angle, any two of its 

 faces together are greater than the third face. It 

 the three angles ASC, BSC, ASB are equal, the 

 proposition is evident. Let ASC be the greatest 

 of the three faces, it is 

 sufficient to prove that 



ASB + BSC ^ ASC. 



At the point S, in the 

 plane ASC, make the 

 angle CSD equal to the 

 angle CSB. Draw any 

 line ADC cutting SA, 

 SD, SC in A, D, and C. 

 Make SD = SB. Join 

 AB, BC. Then in the two triangles CSD, CSB 

 there are two sides CS, SD in the one equal to 

 CS, SB in the other, and the contained angles 

 equal ; therefore CB = CD, and AD is the differ- 

 ence between AC and BC, therefore it is less than 

 AB (Prop. VI., p. 612). Again, in the two tri- 

 angles BSA, DSA there are two sides BS, SA in 

 the one equal to DS, SA in the other ; but the base- 

 BA is greater than DA, therefore 

 the angle BSA is greater than DSA. 

 Consequently the two angles ASB, 

 BSC together are greater than the 

 angle ASC. 



Cor. It will be evident that 

 the two trihedral angles SABC, 

 SA'B'C', are superposable, if 

 SABC has two equal faces. 



PROP. XVIII. In a trihedral angle, if two- 

 faces are equal, the dihedral angles opposite to 

 these faces are also equal. 



Let SABC be a trihedral angle, having the face 



