Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/117

 direction from $P$ to $P'$  is the same as that from $(2)$  to $(1);$  negative, if the contrary of one of these conditions hold; positive again, if the contrary of both conditions be true. In other words, the surface is considered positive if we go around the circumference of the figure $LL'P'P$ in the same sense as $(1)\ (2)\ (3);$  negative, if we go in the contrary sense.

If we consider now a finite part of the line from $L$ to $L'$  and denote by $\phi,$  $\phi'$  the values of the angles at the two extremities, then we have

$$\phi' = \phi + \operatorname{Area} LL'P'P,$$

the sign of the area being taken as explained.

Now let us assume further that, from the origin upon the curved surface, infinitely many other shortest lines go out, and denote by $A$ that indefinite angle which the first element, moving counter-clockwise, makes with the first element of the first line; and through the other extremities of the different curved lines let a curved line be drawn, concerning which, first of all, we leave it undecided whether it be a shortest line or not. If we suppose also that those indefinite values, which for the first line were $\phi,$  $\phi',$ be denoted by $\psi,$  $\psi'$  for each of these lines, then $\psi' - \psi$  is capable of being represented in the same manner on the auxiliary sphere by the space $LL'_{1}P'_{1}P.$  Since evidently $\psi = \phi - A,$  the space

$$\begin{aligned} LL'_{1}P'_{1}P'L'L &= \psi' - \psi - \phi' + \phi \\ &= \psi' - \phi' + A \\ &= LL'_{1}L'L + L'L'_{1}P'_{1}P'. \end{aligned}$$

If the bounding line is also a shortest line, and, when prolonged, makes with $LL',$  $LL'_{1}$ the angles $B,$  $B_{1};$  if, further, $\chi,$  $\chi_{1}$  denote the same at the points $L',$  $L'_{1},$  that $\phi$  did at $L$  in the line $LL',$  then we have

$$\begin{aligned} \chi_{1} &= \chi + \operatorname{Area} L'L'_{1}P'_{1}P', \\ \psi' - \phi' + A &= LL'_{1}L'L + \chi_{1} - \chi; \end{aligned}$$

but

$$\begin{aligned} \phi' &= \chi + B, \\ \psi' &= \chi_{1} + B_{1}, \end{aligned}$$

therefore

$$B_{1} - B + A = LL'_{1}L'L.$$

The angles of the triangle $LL'L'_{1}$ evidently are

$$A,\qquad 180^{\circ} - B,\qquad B_{1},$$