As per the given in question , we draw a figure of a ∆ABC,Here , AB² + AC² = BC² and BC = √2AB,AB² + AC² = BC² ⇒ ∠BAC = 90° ⇒ AB² + AC² = 2AB² ⇒ AB² = AC² ⇒ AB = AC ∴ ∠ABC = ∠ACB = 45°
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