What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ? 

Begin with the Law of Cosines to compute $\cot^2\frac{A}{2}$ in terms of the lengths of the sides of the triangle:
$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$
$$\Rightarrow \begin{cases}
\cos^2{\frac{A}{2}}=\frac{1+\cos{A}}{2}=\frac{2bc+b^2+c^2-a^2}{4bc} = \frac{(b+c)^2-a^2}{4bc} \\
\sin^2{\frac{A}{2}}=\frac{1-\cos{A}}{2}=\frac{2bc-(b^2+c^2-a^2)}{4bc} = \frac{a^2-(b-c)^2}{4bc}
\end{cases}$$
$$\Rightarrow \cot^2\frac{A}{2}=\frac{\cos^2{\frac{A}{2}}}{\sin^2{\frac{A}{2}}}=\frac{(b+c)^2-a^2}{a^2-(b-c)^2}$$
Now, work in the condition from the problem, and let Algebra bring you home ...
$$\begin{eqnarray}
\frac{(b+c)^2}{a^2} &=& \frac{(b+c)^2-a^2}{a^2-(b-c)^2} \\
(b+c)^2\left(a^2-(b-c)^2\right)&=& a^2\left((b+c)^2-a^2\right) \\
a^2(b+c)^2-(b+c)^2(b-c)^2 &=& a^2 (b+c)^2-a^4 \\
a^4-(b+c)^2(b-c)^2 &=& 0 \\
a^4-(b^2-c^2)^2 &=& 0 \\
\left(a^2+(b^2-c^2)\right)\left(a^2-(b^2-c^2)\right) &=& 0 \\
\end{eqnarray}$$
$$\Rightarrow \hspace{0.25in} a^2+b^2=c^2 \hspace{0.25in}\text{or}\hspace{0.25in} a^2+c^2=b^2$$
Nice little problem. I suspect there's a more-direct path to the solution, though. (Edit I see a couple such paths had been posted when I was working on my answer! :)
