Solution of $ 2cos\frac{\beta-\gamma}{2}=\frac{b+c}{\sqrt{b^2+c^2-bc}}$ in a triangle

Question

In $\Delta ABC$ $2\cos\frac{\beta-\gamma}{2}=\frac{b+c}{\sqrt{b^2+c^2-bc}}$, then

(A) $\beta+\gamma=\frac{2\pi}{3}$

(B) $\alpha+\gamma=\beta$

(C) $\alpha=\gamma$

(D) $\beta,\alpha,\gamma,$ are in A.P

This is a multiple choice question which has more then one choice. I am not getting the solution. I cannot approach after $\sqrt{\sin^2\beta+\sin^2\gamma-\sin\beta \sin\gamma}=\cos\frac{\alpha}{2}$

Answer 1

$$\dfrac{2\sin\dfrac{B+C}2\cos\dfrac{B-C}2}{\sqrt{\sin^2A+\sin^2B-\sin B\sin C}}=2\cos\dfrac{B-C}2$$

As $\cos\dfrac{B-C}2\ne0,\sin\dfrac{B+C}2=\cdots=\cos\dfrac A2$

$$\sin^2B+\sin^2C-\sin B\sin C=\dfrac{1+\cos A}2$$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

$$1-\cos(B-C)\cos(B+C)-\dfrac{\cos(B-C)-\cos(B+C)}2=\dfrac{1+\cos A}2$$

Use $\cos(B+C)=\cdots=-\cos A$ to find $$(\cos(B-C)-1)(2\cos A-1)=0$$

Answer 2

In the standard notation we obtain: $$2\cos\frac{\beta-\gamma}{2}=2\left(\sqrt{\tfrac{1+\tfrac{a^2+c^2-b^2}{2ac}}{2}}\cdot\sqrt{\tfrac{1+\tfrac{a^2+b^2-c^2}{2ab}}{2}}+\sqrt{\tfrac{1-\tfrac{a^2+c^2-b^2}{2ac}}{2}}\cdot\sqrt{\tfrac{1-\frac{a^2+c^2-b^2}{2ac}}{2}}\right)=$$ $$=\tfrac{1}{2\sqrt{a^2bc}}\left(\sqrt{(a+b+c)^2(a+b-c)(a+c-b)}+\sqrt{(b+c-a)^2(a+b-c)(a+c-b)}\right)=$$ $$=\frac{(b+c)\sqrt{(a+b-c)(a+c-b)}}{\sqrt{a^2bc}}.$$ Id est,$$(a+b-c)(a+c-b)(b^2-bc+c^2)=a^2bc$$ or $$(a^2-(b-c)^2)(b^2-bc+c^2)=a^2bc$$ or $$(b-c)^2(a^2-b^2+bc-c^2)=0,$$ which gives $$b=c$$ or $$a^2=b^2-bc+c^2,$$ which says $$\alpha=60^{\circ}.$$