ABC triangle with $\tan\left(\frac{A}{2}\right)=\frac{a}{b+c}$ I have a triangle ABC and I know that $\tan\left(\frac{A}{2}\right)=\frac{a}{b+c}$, where $a,b,c$ are the sides opposite of the angles $A,B,C$. Then this triangle is:
a. Equilateral
b. Right triangle with $A=\pi/2$
c. Right triangle with $B=\pi/2$ or $C=\pi/2$ (right answer)
d. Acute
e. Obtuse
I tried to write $\frac{a}{\sin(A)}=2R\implies a=2R\sin(A)$ and to replace in initial equation.Same for $b$ and $c$ but I didn't get too far.
 A: Hint: Use that $$\tan(\frac{\alpha}{2})=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$ where $$s=\frac{a+b+c}{2}$$
Using this we get $$-{\frac { \left( {a}^{2}+{b}^{2}-{c}^{2} \right)  \left( {a}^{2}-{b}^{
2}+{c}^{2} \right) }{ \left( a+b+c \right)  \left( a-b-c \right) 
 \left( b+c \right) ^{2}}}
=0$$
A: Note that
$$
\frac{a}{b+c}=\frac{\sin A}{\sin B+\sin C}=\frac{2\sin(\frac12A)\cos(\frac12A)}{2\sin(\frac12(B+C))\cos(\frac12(B-C))}=\frac{\sin(\frac12A)}{\cos(\frac12(B-C))}
$$
So the given condition is equivalent to
$$
\cos\frac{B-C}2=\cos\frac{A}2
$$
or equivalently, $A\pm B\mp C=0$. Together with $A+B+C=\pi$, we see $B=\frac\pi2$ or $C=\frac\pi2$.
A: One way to see that c) is the correct answer is as follows. Draw yor triangle $ABC$. Construct the angle $A/2$ by extending $BA$ until a point $M$ such that $MA=b$. Note that $MA=AC=b$ so, $MAC$ is isosceles, so you have $\angle CMA=A/2$. Look at the triangle $MBC$ You have $\angle CMB=A/2$ with $CB=a$ and $MB=b+c$. So.... If $\angle B=\pi/2$ you certainly have that $\tan( A/2)=a/(b+c)$. 
A "symmetric" construction proves that if $C=\pi/2$, then you also have $\tan A/2=a/(b+c)$.
Of course this just proves the converse of your statement, but in a multiple choice problem this is good enough.
