\[Proof\]
\[\text{A sphere of radius } r \text{ has surface area } 2\tau{r^{2}}.\]

\[\text{If two great circles meet in a lunar angle }\theta, 0 < \theta \leq{\tau}, \text{then the proportion of surface area which is occupied by the lune they create is } \frac{\theta}{\tau}.\]
\[\therefore \text{ We have area of lune with lunar angle } \theta.\]
\[\frac{\theta}{\tau} \times 2\tau{r^{2}} = 2r^{2}\theta\]

\[\text{Denote by } L_{A}, L_{B} \text{ and } L_{C} \text{ the areas of the three lunes with angles } A, B \text{ and } C, \text{ respectively}.\]
\[\text{Also, Denote by } T \text{ the area of triangle } ABC;\]

\[\text{These areas}(L_{A}, L_{B}, L_{C}) \text{ each have an antipodal duplicate.}\]
\[\text{Likewise, } T \text{ also has its antipodal duplicate.}\]
\[\therefore 2L_{A} + 2L_{B} + 2L_{C} = 2r^{2}\tau + 4T\]

\[Proof\]
\[\text{Let's replace } A + B + C - \frac{1}{2}\tau \text{ with } \varepsilon.\]
\[\therefore T = r^{2}\varepsilon\]
\[\varepsilon = \frac{T}{r^{2}}\]

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