$ABC$ is an isosceles triangle right-angled at $C$. Prove that $AB^{2} = 2AC^{2}$.

(N/A) Given that $\triangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.
Since it is an isosceles triangle,the two legs must be equal,so $AC = CB$.
Applying the Pythagoras theorem in $\triangle ABC$ (which is right-angled at point $C$),we have:
$AC^{2} + CB^{2} = AB^{2}$
Since $AC = CB$,we can substitute $CB$ with $AC$ in the equation:
$AC^{2} + AC^{2} = AB^{2}$
$2AC^{2} = AB^{2}$
Hence,it is proved that $AB^{2} = 2AC^{2}$.