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!

HINT:
$$F=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$
Now using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=\cos(A+B)\cos(A-B)+\cos^2C+1$$
Now as $\cos(A+B)=\cos(\pi-C)=-\cos C,$
$$F=-\cos C\cos(A-B)+\cos C\{-\cos(A+B)\}=1-\cos C(2\cos A\cos B)$$
