How to divide a spherical triangle into two equal-area spherical triangles? 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.
 A: 
Consider the spherical triangle $ \mathscr A$$ \mathscr B $$ \mathscr C $depicted in the figure, with angular length of its sides a, b, c and angle subtended at the corners A, B, C. Then, the solid angle (or its surface area in case of sphere of unit radius) covered is $E=A+B+C-π$
Let the point $ \mathscr D $ on $ \mathscr B$$ \mathscr C$ be such that the geodesic $ \mathscr A$$ \mathscr D $ divides $ \mathscr A$$ \mathscr B$$ \mathscr C$ into two smaller spherical triangles of equal area. Now, we shall solve for the angle x marked in the figure.
Area of spherical triangle $ \mathscr A$$ \mathscr B $$ \mathscr D $ is: $ x+y+b-π=E/2=(A+B+C-π)/2$
$$y=\frac{A+C-B+π}{2}-x$$
$$cos(y)=cos(\frac{A+C-B+π}{2})cos(x)+sin(\frac{A+C-B+π}{2})sin(x)$$
In $ \mathscr A$$ \mathscr B $$ \mathscr D $, the spherical trigonometry relation between c and included angles is,
$$cos(y)=cos(c)sin(B)sin(x)-cos(B)cos(x) $$
Eliminating cos(y) in the above two equations, the relation for calculating angle x for the given triangle $ \mathscr A$$ \mathscr B $$ \mathscr C $ can be got,
$$tan(x)=\frac{cos(B)-sin(\frac{A+C-B}{2})}{sin(B)cos(c)-cos(\frac{A+C-B}{2})} $$
In a similar fashion, $ \mathscr A$$ \mathscr B$$ \mathscr C$ could be divided into two equal area spherical triangles in two more ways using suitable points (say $ \mathscr E$ on $ \mathscr C$$ \mathscr A$  and $ \mathscr F$ on $ \mathscr A$$ \mathscr B$). The three geodesic arcs $ \mathscr A$$ \mathscr D$, $ \mathscr B$$ \mathscr E$ and $ \mathscr C$$ \mathscr F$ meet at a common point (say, $ \mathscr H$) within spherical triangle $ \mathscr A$$ \mathscr B$$ \mathscr C$. However, the three sub-triangles $ \mathscr A$$ \mathscr B$$ \mathscr H$, $ \mathscr B$$ \mathscr C$$ \mathscr H$ and $ \mathscr C$$ \mathscr A$$ \mathscr H$ do not generally exhibit equal area.
A: Gauss Bonnet theorem for polygons on sphere (radius $a$) surface has ( $ K = 1/a^2 $) 
$$ 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.
