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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.

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    $\begingroup$ The area of a spherical triangle is the spherical excess: en.wikipedia.org/wiki/… - that should tell you how to divine one angle to make the two parts equal. $\endgroup$ – Ethan Bolker Apr 5 '16 at 14:45
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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.

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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.

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