Side of Spherical Triangle is Supplement of Angle of Polar Triangle

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Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.

Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.

Then $A'$ is the supplement of $a$.

That is:

$A' = \pi - a$

and it follows by symmetry that:

$B' = \pi - b$
$C' = \pi - c$



Let $BC$ be produced to meet $A'B'$ and $A'C'$ at $L$ and $M$ respectively.

Because $A'$ is the pole of the great circle $LBCM$, the spherical angle $A'$ equals the side of the spherical triangle $A'LM$.

That is:

$(1): \quad \sphericalangle A' = LM$

From Spherical Triangle is Polar Triangle of its Polar Triangle, $\triangle ABC'$ is also the polar triangle of $\triangle A'B'C'$.

That is, $C$ is a pole of the great circle $A'LB'$.

Hence $CL$ is a right angle.

Similarly, $BM$ is also a right angle.

Thus we have:

\(\ds LM\) \(=\) \(\ds LB + BM\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds LB + \Box\) where $\Box$ denotes a right angle

By definition, we have that:

$BC = a$
\(\ds BC\) \(=\) \(\ds a\) by definition of $\triangle ABC$
\(\ds \leadsto \ \ \) \(\ds LB + a\) \(=\) \(\ds LC\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds LB\) \(=\) \(\ds \Box - a\) as $LC = \Box$


\(\ds \sphericalangle A'\) \(=\) \(\ds LM\) from $(1)$
\(\ds \) \(=\) \(\ds LB + \Box\) from $(2)$
\(\ds \) \(=\) \(\ds \paren {\Box - a} + \Box\) from $(3)$
\(\ds \) \(=\) \(\ds \paren {2 \Box} - a\)

where $2 \Box$ is $2$ right angles, that is, $\pi$ radians.

That is, $A'$ is the supplement of $a$:

$A' = \pi - a$

By applying the same analysis to $B'$ and $C'$, it follows similarly that:

$B' = \pi - b$
$C' = \pi - c$