# Side of Spherical Triangle is Supplement of Angle of Polar Triangle

## Theorem

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:

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

## Proof 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 $LC = \Box$

Then:

 $\ds \sphericalangle A'$ $=$ $\ds LM$ 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:

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

$\blacksquare$