# 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.

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 $$-{\frac { \left( {a}^{2}+{b}^{2}-{c}^{2} \right) \left( {a}^{2}-{b}^{ 2}+{c}^{2} \right) }{ \left( a+b+c \right) \left( a-b-c \right) \left( b+c \right) ^{2}}} =0$$
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 $$C=\frac\pi2$$.
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 $$\tan A/2=a/(b+c)$$.