If $2R+r=r_1$, then prove that $A=90$ 
R is the circumradius of the triangle ABC, $r$ is the inradius and $r_1$ is the ex-radius on side a which is opposite to angle A

In the expression 
$$2R+4R\sin \frac A2 \sin \frac B2 \sin \frac C2=4R\sin \frac A2 \cos \frac B2 \cos \frac C2$$
R gets canceled and the expression becomes 
$$2+\sin \frac A2(\cos \frac{B-C}{2}-\cos \frac{B+C}{2})=\sin \frac A2 (\cos \frac{B+C}{2}+\cos \frac{B-C}{2})$$
$$2-\sin^2\frac A2=\sin ^2\frac A2$$
$$\sin \frac A2=1$$
$$A=180$$ which is obviously wrong. Please let me know my mistake.

Hint
$$2R=\cdots=4R\sin A/2\cos(B+C)/2=4R\sin^2A/2=2R(1-\cos A)$$
