Answer: B
In triangle ABC, AB\(^2\) + BC\(^2\) = AC\(^2\) ∴ The triangle is a right-angled triangle Again, AC = \(\sqrt{2}\) × BC ∴ AB\(^2\) + BC\(^2\) = 2BC\(^2\) Or, AB\(^2\) = BC\(^2\) Or, AB = BC ∴ The triangle is a right-angled isosceles triangle
In triangle ABC, AB\(^2\) + BC\(^2\) = AC\(^2\) ∴ The triangle is a right-angled triangle Again, AC = \(\sqrt{2}\) × BC ∴ AB\(^2\) + BC\(^2\) = 2BC\(^2\) Or, AB\(^2\) = BC\(^2\) Or, AB = BC ∴ The triangle is a right-angled isosceles triangle