= Solution
Use <suffix notation>, summing repeated indices from one to three. The contraction identity for the <Levi-Civita symbol> gives
$$
\begin{aligned}
(a\times b)\cdot(a\times c)
&=\epsilon_{ijk}\epsilon_{i\ell m}a_jb_ka_\ell c_m\\
&=(\delta_{j\ell}\delta_{km}-\delta_{jm}\delta_{k\ell})a_jb_ka_\ell c_m\\
&=(a\cdot a)(b\cdot c)-(a\cdot c)(b\cdot a)\\
&=b\cdot c-(a\cdot b)(a\cdot c).
\end{aligned}
$$
Here $a\cdot a=1$ because $a$ is a <unit vector>. For the second <cross product> identity, use $\epsilon_{ijk}\epsilon_{j\ell m}=\delta_{im}\delta_{k\ell}-\delta_{i\ell}\delta_{km}$:
$$
\begin{aligned}
[(a\times b)\times(a\times c)]_i
&=\epsilon_{ijk}\epsilon_{j\ell m}\epsilon_{knp}a_\ell b_ma_nc_p\\
&=b_i\epsilon_{\ell np}a_\ell a_nc_p-a_i\epsilon_{mnp}b_ma_nc_p\\
&=0-a_i\,b\cdot(a\times c)\\
&=[a\cdot(b\times c)]a_i.
\end{aligned}
$$
The first term vanishes because $a_\ell a_n$ is symmetric while the <Levi-Civita symbol> is antisymmetric in those indices. The last equality uses cyclicity and antisymmetry of the <scalar triple product>.
On the <unit sphere>, a <great circle> has radius one, so the <spherical distance> is its shorter arc length, equal to its central angle in $[0,\pi]$. The <dot product> of its endpoint <unit vectors> is the cosine of that angle. Hence
$$
\boxed{\cos\delta(A,B)=a\cdot b.}
$$
For the angular formulas, take the ordinary nondegenerate convex minor <spherical triangle>: no pair of vertices is antipodal and the three vertices do not lie on one <great circle>. These are needed for the displayed denominators and interior angles to be defined. Merely requiring three distinct points, as the PDF does, does not exclude degeneracy.
Put $p=\delta(A,B)$, $q=\delta(A,C)$ and $r=\delta(B,C)$. The outgoing unit tangent <vectors> at $A$ are
$$
u=\frac{b-(a\cdot b)a}{\sin p},\qquad v=\frac{c-(a\cdot c)a}{\sin q}.
$$
Their <dot product> is $\cos\alpha$. Equivalently, they are obtained from the normalized plane normals $a\times b$ and $a\times c$ by the same right-angle rotation in the tangent plane: $u=(a\times b)\times a/|a\times b|$, and similarly for $v$. Rotation preserves the <dot product>, giving
$$
\cos\alpha=\frac{(a\times b)\cdot(a\times c)}{|a\times b|\,|a\times c|}.
$$
Since $|a\times b|=\sin p$ and $|a\times c|=\sin q$, the first proved identity becomes the <spherical law of cosines>:
$$
\boxed{\cos r=\cos p\cos q+\sin p\sin q\cos\alpha.}
$$
For the <spherical law of sines>, take norms in the second proved identity. Writing $\tau=a\cdot(b\times c)$, and using $0<\alpha<\pi$, gives
$$
|\tau|=\sin p\sin q\sin\alpha.
$$
Cyclic permutation gives the corresponding formulas at $B$ and $C$. Dividing each by the positive product $\sin p\sin q\sin r$ proves
$$
\boxed{\frac{\sin\alpha}{\sin r}=\frac{\sin\beta}{\sin q}=\frac{\sin\gamma}{\sin p}=\frac{|\tau|}{\sin p\sin q\sin r}.}
$$
Finally, if all sides equal $s$, the <spherical law of cosines> gives
$$
\cos\alpha=\frac{\cos s-\cos^2s}{\sin^2s}=\frac{\cos s}{1+\cos s}<\frac12,
$$
because $0<s<\pi$ makes $1+\cos s>0$ and $\cos s<1$. Cosine is strictly decreasing on $(0,\pi)$, so \b[every angle of a nondegenerate equilateral <spherical triangle> is greater than $\pi/3$].
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