Proof that a spherical triangle is congruent with its dual I'm currently studying spherical geometry and ran into an exercise problem that I'm having trouble understanding.
The book first defines a dual spherical triangle $\triangle^* ABC$ of original $\triangle ABC$ to be defined by the three points:
$$
\begin{align}
A^* & = \dfrac{B \times C}{\sin{(a)}} \\
B^* & = \dfrac{C \times A}{\sin{(b)}} \\
C^* & = \dfrac{A \times B}{\sin{(c)}}
\end{align}
$$
If the opposite sides of each point were called $a^*$, $b^*$, $c^*$, then the following theorem holds:

$\triangle ABC$ and $\triangle^* ABC$ are congruent, and therefore:
$$A = \pm \dfrac{B^* \times C^*}{\sin{(a^*)}} \quad B = \pm \dfrac{C^* \times A^*}{\sin{(b^*)}} \quad C = \pm \dfrac{A^* \times B^*}{\sin{(C^*)}}$$

Proof:

We prove that $B^* \times C^*$ and $A$ are parallel (i.e. $B^* \times C^* \parallel (C \times A) \times (A \times B)$) by using the vector triple product (a.k.a. Lagrange's formula):
$$(C \times A) \times (A \times B) = \langle C \times A, B \rangle A - \langle C \times A, A \rangle B = \text{det}(A, B, C) A$$
which implies that the vector is parallel to $A$ and we can write:
$$A = \pm \dfrac{B^* \times C^*}{\sin{(a^*)}}$$

The two main parts that I'm having trouble with is understanding the theorem of congruence itself and the proof.

*

*How does $A = \pm \frac{B^* \times C^*}{\sin(a^*)}$ imply congruence? Where did it even come from?

*In the proof, how did the calculation go from the triple product to the determinant (I assume)? And is proving $(C \times A) \times (A \times B) \parallel A$ enough to write $A = \pm \frac{B^* \times C^*}{\sin(a^*)}$?

Thanks in advance.
 A: First of all, note that your definition of $B^*$ is wrong: One has to write $B^* = \frac{C \times A}{\sin(b)}$ because in the end, one would like to end up with three times the same sign in the statement of the Theorem (i.e., $A = \pm A^*$, $B = \pm B^*$, $C = \pm C^*$, where all the signs are equal).
Now, that $\Delta ABC$ and $\Delta^* ABC$ are congruent is not correct. In fact, what the Theorem says is that the dual triangle of the dual triangle is the original triangle (or the antipodal one, if one has a negative sign). So $\Delta ABC$ and $\Delta^{**} ABC$ are congruent (since their sets of side-lengths coincide).
To see that $\Delta ABC$ and $\Delta^* ABC$ are not always congruent, denote by $\alpha, \beta, \gamma$ the angles of $\Delta ABC$ (with the standard notation), then one has
$$a^* = \sphericalangle(C \times A, A \times B) = \sphericalangle(-A \times C, A \times B) = \pi - \sphericalangle(A \times C, A \times B) = \pi - \alpha.$$
To explain the second to last equality sign, since $-\cos(a) = \cos(\pi - a)$ for any real number $a$, we obtain
$$\cos(\pi - \sphericalangle (x,y)) = - \cos\sphericalangle(x,y) = -\frac{\langle x,y\rangle}{||x|| \cdot ||y||} = \frac{\langle -x,y\rangle}{||-x|| \cdot ||y||} = \cos\sphericalangle (-x,y)$$
for $x,y \in \mathbb R^3$.
In the same way, we obtain that $b^* = \pi - \beta$ and $c^* = \pi - \gamma$. Now, it is not difficult to find examples of triangles such that $\{a,b,c\} \neq \{a^*,b^*,c^*\}$; just play around a little bit to find some. Since the sets of side-lengths of congruent triangles must be equal, we see that $\Delta ABC$ and $\Delta^* ABC$ are not always congruent.
To your second question, one can verify the formula $$(x \times y) \times (w \times z) = z \cdot \det(x,y,w) - w \cdot \det(x,y,z)$$ for $x,y,z,w \in \mathbb R^3$ via a direct computation. It is a little bit tedious, but a good and easy exercise.
Finally, if one has $(C \times A) \times (A \times B) \parallel A$, one has $B^* \times C^* \parallel A$ by definition of $B^*$ and $C^*$. Since both $A$ and
$$\frac{B^* \times C^*}{||B^* \times C^*||} = \frac{B^* \times C^*}{\sin(a^*)}$$
lie on the unit sphere in $\mathbb R^3$, the only possibilities for them to be congruent are $A = \pm \frac{B^* \times C^*}{\sin(a^*)}$.
