Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!

$$\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$$
Use $\cos^2{A} =\frac{1+\cos2A}2$. Then 
$$\cos2A+\cos2B+\cos2C=-1$$
Use $\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$
Then $$-1-4\cos A\cos B\cos C=-1$$
$$4\cos A\cos B\cos C=0$$
Then $\angle A$ or $\angle B$ or $\angle C$ is $\frac{\pi}{2}$
