Proving the excess of two split triangles is the same as the original

noneuclidean-geometrytriangles

The excess of a triangle is the number of degrees over $180$ a triangle has in spherical geometry.

Given spherical $\triangle ABC$ that is split into two triangles by an arc from the vertex going to the opposite side. Show that the excess of $\triangle ABC=$ the excess of $\triangle ABD$+ the excess of $\triangle BDC$.

I drew this picture: enter image description here

Essentially it seems that $\angle ABC=\angle ABD+\angle DB$C.

I therefore need the excess defined by the two "bottom angles" of each smaller triangle to be the same as the excess defined by $\angle BAC$ and $\angle BCD$.
Because each of the smaller triangles only has one angle that is different than those of the larger triangle, the excess from the two angles formed by $d$ needs to be half of what is contributed by $\angle BAD$ and $\angle BCD$. Is this true?

Best Answer

Let

  • $\alpha$, $\beta$, $\gamma$ be the interior angles of the larger triangle at $A$, $B$, $C$, respectively,
  • $\beta_1$, $\beta_2$ be the angles of the smaller triangles at $B$, and
  • $\delta_1$, $\delta_2$ be the angles of the smaller triangles at $D$.

Then $\beta_1 + \beta_2 = \beta$ and $\delta_1 + \delta_2 = 180$ degrees. Therefore the sum of excesses of the smaller triangles is $$ (\alpha + \beta_1 + \delta_1 - 180) + (\gamma+\delta_2 + \beta_2 - 180) = \alpha + \beta + \gamma - 180 $$ and therefore equal to the excess of the larger triangle.

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