How to divide a spherical triangle into two equal-area spherical triangles?

Question

Median of a planar triangle divides it into two equal-area smaller triangles. Whereas, in a spherical triangle, the geodesic joining the corner to the midpoint of the opposite side does not divide the spherical triangle into two equal-area parts! I have searched several textbooks on 'Spherical Trignometry' including the classic book by I. Todhunter. But, a geometrical method or a formula for implementing the exact division of spherical triangle into two equal-area spherical triangles seems to be unavailable.

Answer 1

Gauss Bonnet theorem for polygons on sphere (radius $a$) surface has ( $ K = 1/a^2 $)

$$ KA = \Sigma \alpha_i - \pi $$

for a spherical triangle and, similarly for a smaller triangle of half such area

$$ KA/2 = \Sigma \beta_i - \pi $$

taking difference

$$ KA/2 = \Sigma(\alpha_i - \beta_i) $$

So at each vertex this much extra rotation at external angle of produced geodesic side of spherical polygon defines a slimmer triangle. There are a number of ways this can be implemented, there is no unique way. The procedure is general, this way we can bifurcate an icosahedron area also, for example.