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{\bf Question}

Let $T$ be a triangle in ${\bf H}$ with vertices $v_a$, $v_b$, and
$v_c$.  Show that the ray from $v_a$ bisecting the angle at $v_a$
contains the midpoint of the hyperbolic line segment $\ell$
joining $v_b$ and $v_c$ if and only if the angles at $v_b$ and
$v_c$ are equal.
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{\bf Answer}

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Suppose $\beta=\gamma$. Then by lcII, using side of length $d$
implies that $\phi=\mu$. Since $\beta=\gamma$ and $\phi=\mu$,
applying lcII to both subtriangle yields that  \begin{eqnarray*}
\cosh(a') & = & \cosh(a'')\\ & = &
\ds\frac{\cos(\theta)+\cos(\gamma)\cos(\mu)}{\sin(\gamma)\sin(\mu)}
\end{eqnarray*} and so $a'=a''$.

Suppose now that $a'=a''$. Then, using ls, we see that

$\ds\frac{\sinh(a')}{\sin(\theta)}=\ds\frac{\sinh(d)}{\sin(\beta)}=
\ds\frac{\sinh(d)}{\sin(\gamma)}=\ds\frac{\sinh(a'')}{\sin(\theta)}$
and so $\beta=\gamma$, as desired.
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