In $\Delta ABC$, if $2\cos\frac{A-C}{2}=\frac{a+c}{\sqrt{a^2+c^2-ac}}$, then choose the right options

Question

1. $B=\frac{\pi}{3}$
2. $B=C$
3. $A, B, C$ are in arithmetic progression
4. $B+C=A$

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

$$\sqrt {\sin^2A+\sin^2C-\sin A\sin C}=\sin \frac{A+C}{2}$$

$$\sin^2A+\sin^2C-\sin A\sin C = \cos ^2 \frac B2$$

What should I do next?

Answer 1

Continue with

\begin{align} & \sin^2A+\sin^2C-\sin A\sin C - \cos ^2 \frac B2 \\ =&(\sin A - \sin C)^2 + \frac12( \cos (A-C)-\cos (A+C))- \frac12(1+ \cos B) \\ =& 4\cos^2 \frac{A+C}2 \sin^2 \frac{A-C}2 - \frac12(1- \cos (A-C)) \\ =& 4\sin^2 \frac{B}2 \sin^2 \frac{A-C}2 - \sin^2 \frac{A-C}2 \\ = &(4\sin^2 \frac{B}2 -1)\sin^2 \frac{A-C}2 \\ \end{align}

which leads to $\sin \frac B2 = \frac12 $ or $\sin\frac{A-C}2 =0$. Thus, $ B= \frac\pi3$ or $A =C$

Answer 2

Use Law of Sines

$$\dfrac{c+a}{2\cos\dfrac{C-A}2}=\cdot=2R\cos\dfrac B2>0$$

$$\implies\left(2R\cos\dfrac B2\right)^2=4R^2(\sin^2C+\sin^2A-\sin C\sin A)$$

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

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$1+\cos B=2-2\cos(C-A)\cos (C+A)-[\cos(C-A)-\cos(C+A)]$$

Finally, using $\cos(C+A)=\cdots=-\cos B$

We get $$(\cos(C-A)-1)(2\cos B-1)=0$$

Can you take it from here?

Answer 3

In the standard notation we obtain: $$4\cos^2\frac{\alpha-\gamma}{2}=\frac{(a+c)^2}{a^2-ac+c^2}$$ or $$1+\cos(\alpha-\gamma)=\frac{(a+c)^2}{2(a^2-ac+c^2)}$$ or $$\cos\alpha\cos\gamma+\sin\alpha\sin\gamma=\frac{-a^2+4ac-c^2}{2(a^2-ac+c^2)}$$ or $$\frac{b^2+c^2-a^2}{2bc}\cdot\frac{a^2+b^2-c^2}{2ab}+\frac{2S}{bc}\cdot\frac{2S}{ab}=\frac{-a^2+4ac-c^2}{2(a^2-ac+c^2)}$$ or $$\frac{b^4-(a^2-c^2)^2+\sum\limits_{cyc}(2a^2b^2-a^4)}{2b^2ac}=\frac{-a^2+4ac-c^2}{a^2-ac+c^2}$$ or $$\frac{b^2(a^2+c^2)-(a^2-c^2)^2}{2b^2ac}=\frac{-a^2+4ac-c^2}{a^2-ac+c^2}$$ or $$(a^2-c^2)^2(a^2+c^2-ac-b^2)=0,$$ which gives $$a=c$$ or $$\beta=60^{\circ}.$$ We see that any option may be wrong.