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 ? 
 A: So by Law of Sines we have $$ \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} =k (\text{say})$$
From this your equation becomes, $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{k(\sin{B} + \sin{C})}{k \sin{A}} = \frac{\sin{B}+\sin{C}}{\sin{A}} = \frac{\cos\frac{A}{2} \cos\frac{B-C}{2}}{\sin\frac{A}{2} \cos\frac{A}{2}}$$ From this you get $$\frac{\pi}{2} - \Bigl(\frac{B+C}{2}\Bigr)= \pm \frac{B-C}{2}$$ and you can do it then.
This gives $B=\frac{\pi}{2}$ or $C =\frac{\pi}{2}$, which means the triangle is right-angled. Hurray!
NOTE: The keen idea when you see problems of this type is to use the Sine rule. Do you think that this problem would have been difficult, if you had applied the sine rule as i did in the first step. Very often in Mathematics starting the solution is the difficult part. Once you figure that out, then the solutions simply flows. 
A: The cosine formula states
$$a^2 = b^2+ c^2 – 2bc \cos A$$
and since
$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$
we have
$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$
which reduces to $a^4 = (b^2 – c^2)^2.$
And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.
Hence $b^2 = c^2+a^2$ and so we have a right-angled triangle.
And so it's the angle opposite the larger of $b$ and $c$ which is the right-angle.
A: OK, first rewrite it as
$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$
Multiply both sides by $2 \cos(A/2)$ 
$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$
Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:
$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$
Using Simpson's rule on the second hand of the equation to sum the sines gives:
$$ 2 \cos^2(A/2) = 2\sin((B+C)/2)\cos((B-C)/2) $$
Since in a triangle $A+B+C=\pi$ we have $ 2\sin((B+C)/2) = 2\sin((\pi-A)/2) = 2\cos(A/2) $ and thus
$$ \cos(A/2) = \cos((B-C)/2) $$
Or $A+C=B$ (If $B>C$). Combining this with $A+B+C=\pi$ this gives $B=\pi/2$.
A: 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! :)
A: I tried a geometry proof for this and thought it was worth sharing it here.
I am unable to draw the figure. I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures.
Consider the triangle $ABC$. Let the angle bisector of $A$ meet $BC$ at $D$.
By angle bisector theorem, we get
$\frac{AB}{AC} = \frac{DC}{DB}$. Hence, we get $\frac{b}{c} = \frac{DC}{DB}$.
Add one to both sides, we get
$\frac{b+c}{c} = \frac{DC+DB}{DB} = \frac{a}{DB}$.
Rearranging we get $\frac{c}{DB} = \frac{b+c}{a} = \cot(\frac{A}{2})$.
A similar argument (or using the fact that $\frac{c}{DB} = \frac{b}{DC}$) yields, $\frac{b}{DC} = \frac{b+c}{a} = \cot(\frac{A}{2})$.
Now, we are almost done. (Note that if $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$, the above is true.) However, we still need to prove that for no other possibility this is satisfied.
To prove the above. Draw a perpendicular to $AB$ at $B$ and similarly draw a perpendicular to $AC$ at $C$. Let these two will intersect $AD$ at $D'$ and $D''$. Now note that $BD' = BD$ since $\frac{BD}{AB} = \cot(\frac{A}{2}) = \frac{BD'}{AB}$ (Since $D'BA$ is a right angled triangle). Similarly $CD'' = CD$.
But this is not possible unless $D' = D$ or $D'' = D$, which implies $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$
Again, I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures. The argument I hope however gives you the idea behind the method.
