Prove that for any triangle: $(a-b) \cos \hat{C} = c(\cos\hat{B}-\cos \hat{A}) \Rightarrow a = b , \hat{A} = \hat{B}$

Question

I'm working on this problem (which must be very easy but for some reason I can't see the solution): Suppose that you know that $$(a-b) \cos \hat{C} = c(\cos\hat{B}-\cos \hat{A}),$$ now prove that $a=b$ and that $ \hat{A}=\hat{B}$. I'm inclined to use the law of cosines: $$a^2 = b^2 + c^2 - 2bc \cos \hat{A}$$ $$b^2 = a^2 + c^2 - 2ac \cos \hat{B}$$ $$c^2 = a^2 + b^2 - 2bc \cos \hat{C}$$ but that didn't lead me anywhere in particular. Any leads/hints/solutions/ideas?

Answer 1

Hint: Use the Theorem of cosines: $$(a-b)\left(\frac{a^2+b^2-c^2}{2ab}\right)=c\left(\frac{a^2+c^2-b^2}{2ac}-\frac{b^2+c^2-a^2}{2bc}\right)$$ and the right-Hand side simplifies to $$c\left(1/2\,{\frac { \left( a-b \right) \left( a+b+c \right) \left( a+b-c \right) }{acb}} \right)$$ Can you finish?

Answer 2

Hint: $$a=b\cos\hat{C}+c\cos\hat{A}$$ (for a=b)

And use: $$a=2R\sin\hat{A},b=2R\sin\hat{B}$$ in previous result for $\hat{A}=\hat{B}$