Find the nature of the triangle Consider a triangle $ABC$ for which $\tan \frac{A}{2} = \frac{a}{b + c}$, where $a = BC, \: b = AC, \:c = AB$.
Then, the triangle is:

A) equilateral
B) right triangle with $A = \frac{\pi}{2}$
C) right triangle with $B = \frac{\pi}{2}$ or $C = \frac{\pi}{2}$
D) acute
E) obtuse

All I could do was to eliminate "$A$", which is pretty obvious.

Option C) is correct by  trig simplification.
By Sine Rule $a$ is proportional to  $\sin A $ with circumcircle diameter as constant of proportionality.
Using shorthand for $ ( s= \sin, c=\cos )$
$$ \tan A/2 = \dfrac{s_{A/2}}{c_{A/2}}= \dfrac{s_A}{s_B+s_C}= \dfrac{2 s_{A/2}c_{A/2}}{2 s_{(B+C)/2}c_{(B-C)/2}} \tag1 $$
since 
$$ \frac{(A+B+C)}{2}=\pi/2,\tag2 $$
$$ \cos A/2 = \sin (B+C)/2  \tag3 $$
Plug into 1) RHS, it simplifies to
$$ \dfrac{s_{A/2}}{c_{(B-C)/2}} \tag4 $$
LHS
$$ \tan A/2 = \dfrac{s_{A/2}}{c_{A/2}} \tag5 $$
In RHS of 4) and 5) numerators are same, so equate denominators
$$ \cos A/2 = \cos \frac{B-C}{2} \tag6 $$
For inverse cosine function equality we should have either
$$ \frac{A}{2}  = + \frac{B-C}{2} \tag7$$
or
$$ \frac{A}{2} = - \frac{B-C}{2} \tag8  $$
And further on comes to either $B= \pi/2 $ or $C= \pi/2 $ making it a right triangle.
