Show that $r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$

Question

In triangle $\Delta~ ABC$, $~r~$ is the in-radius and $~[ABC]~$ is the area.

Please explain $$ r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$$ thanks.

Answer 1

$$\cot(A/2)=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}~ \mbox{etc} ~\mbox{and}~ \Delta=\sqrt{s(s-a)(s-b)(s-c)}.$$ $\Delta$ denotes thr area $[ABC]$ and $s$ is semiparameter. So the given expression (call it $F$), is $$F=r^2s \frac{\sqrt{s(s-a)(s-b)(s-c)}}{(s-a)(s-b)(s-c)}=\frac{r^2s^2\Delta}{\Delta^2}=\frac{\Delta^3}{\Delta^2}=\Delta.$$ Here $\Delta$ is area and we have used $\Delta=rs.$.

Answer 2

\begin{align} [ABC]&=[IAB]+[IBC]+[ICA] . \end{align}

\begin{align} [IAB]&=\tfrac12\cdot|AB|\cdot|IC_t| =\tfrac12(|AC_t|+|BC_t|)\cdot|IC_t| =\tfrac12(r\cot\tfrac\alpha2+r\cot\tfrac\beta2)\cdot r =r^2(\tfrac12\,\cot\tfrac\alpha2+\tfrac12\,\cot\tfrac\beta2) . \end{align}

Similarly,

\begin{align} [IBC]&= r^2(\tfrac12\,\cot\tfrac\beta2+\tfrac12\,\cot\tfrac\gamma2) ,\\ [ICA]&= r^2(\tfrac12\,\cot\tfrac\gamma2+\tfrac12\,\cot\tfrac\alpha2) . \end{align}

Hence,

\begin{align} [ABC]&=[IAB]+[IBC]+[ICA] =r^2(\cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2) . \end{align}

And \begin{align} \cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2 &= \cot\tfrac\alpha2\cot\tfrac\beta2\cot\tfrac\gamma2 \quad \text{ for }\alpha+\beta+\gamma=180^\circ \end{align}
is a known triple cotangent identity, which is easy to check using substitution

\begin{align} \cot\tfrac\gamma2&= \frac{\cot\tfrac\alpha2+\cot\tfrac\beta2}{\cot\tfrac\alpha2\cot\tfrac\beta2-1} . \end{align}