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 ? 

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.
