How to prove $AC^2 + BC^2 = 2 AB^2$, if $\cot A + \cot B = 2 \cot C$ in $\triangle ABC$?

Question

Prove, for $\triangle ABC$, that $$AC^2 + BC^2 = 2\,AB^2$$ if $$\cot A + \cot B = 2 \cot C$$

Can this be solved by the sine rule?

$\cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin A \cos B + \sin B \cos A}{\sin A \sin B}$

$= \frac{\sin(A+B)}{\sin A \sin B} = 2 \frac{\cos C}{\sin C}$

But because $A+B = \pi - C$ so $\sin(A+B) = \sin C$

Therefore:

$\frac{\sin C}{\sin A \sin B} = 2 \frac{\cos C}{\sin C}$ $\sin^2 C = 2 \sin A \sin B \cos C$

after that, what I have to do?

Answer 1

Hint:

We can use Sine & Cosine Law:

$$\cot A+\cot B=2\cot C$$

$$\iff\dfrac{b^2+c^2-a^2+c^2+a^2-b^2}{2abc}=2\left(\dfrac{a^2+b^2-c^2}{2abc}\right)$$