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{\bf Question}

Show that a triangle $T$ in ${\bf H}$ is an equilateral triangle
(that is, all sides have the same length) if and only if its
angles are all equal.

\medskip
\noindent Now, let $\alpha$ be the angle at a vertex of $T$ and
let $a$ be the hyperbolic length of a side of $T$.  Show that
$2\cosh(\frac{1}{2} a) \sin(\frac{1}{2}\alpha) = 1$.

\medskip

{\bf Answer}

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\un{by lcI}: all angles are equal and have
$\cos(\alpha)=\ds\frac{\cosh^2(a)-\cosh(a)}{\sinh^2(a)}$ (and the
fact that an angle in the range $[0,\pi]$ is completely determined
by its cosine)

(the converse, that equal angles imply equal side lengths follows
immediately from lcII, with

$$\cosh(a)=\ds\frac{\cos(\alpha)-\cos^2(\alpha)}{\sin^2(\alpha)})$$

The bisector of the angles intersects the opposite side in a right
angle by a geometric argument, namely the triangle is taken to
itself by reflection in the bisecting line.  The same argument
shows that the bisecting line intersects the opposite side in its
midpoint.

Now use \un{lcII}.
$\cos(\frac{1}{2}\alpha)=-\cos(\alpha)\cos(\frac{\pi}{2})+
\sin(\alpha)\sin(\frac{\pi}{2})\cosh(\frac{1}{2}a)$

$\cosh(\frac{1}{2}a)=\ds\frac{\cos(\frac{1}{2}\alpha)}{\sin(\alpha)}=
\ds\frac{1}{2\sin(\frac{1}{2}\alpha)}$

(since
$\sin(\alpha)=2\sin(\frac{1}{2}\alpha)\cos(\frac{1}{2}\alpha))$

and so $2\cosh(\frac{1}{2}a)\sin(\frac{1}{2}\alpha)=1$ as desired.
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