Stewart theorem validity on a sphere

Question

EDIT1:

On a spherical surface radius $R$ a geodesic triangle is drawn:

Let a,b, and c be the lengths of the sides of the geodesic triangle. Let d be the geodesic arc length of a cevian to the side of length a. The cevian divides the side of length a into two geodesic arc segments of length m and n, with m adjacent to c and n adjacent to b.

Would a * spherical Stewart's theorem*

$${\displaystyle b^{2}m+c^{2}n=a(d^{2}+mn)}? $$

be still valid? I do not think so.

If not, what changes in the angles of spherical triangle be incorporated into a new relation?

The radius of sphere is thrice the circum-diameter of a plane triangle with these dimensions.

It is derived using cosine law in the plane. So spherical trigonometry is valid it appears. But wanted to find more applicable grounds... — Narasimham, May 29 '20 at 08:27
@ ArcticChar I have edited so that we might see how the relations would change at least .on a spherical triangle. — Narasimham, May 29 '20 at 19:46

K.Jeong · Answer 1 · 2020-08-12T10:57:40.650

2

We have $\sin a \cos d = \sin m \cos b + \sin n \cos c$ for a spherical triangle. And by third-order approximation, we can recover the original Stewart theorem $a^3 + 3ad^2 = m^3 + 3mb^2 + n^3 + 3nc^2$.

edited Aug 12 '20 at 10:57

answered Aug 11 '20 at 05:29

K.Jeong

21

K.Jeong, @Narasimham: Thanks for having pointed out the applicability of Stewart's theorem. Please note my recent related question "Generalizing Stewart's theorem to timelike triangles in Lorentzian spaces of constant curvature" (mse/q/4029996) – user12262 Feb 18 '21 at 07:15

Stewart theorem validity on a sphere

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