ABC is an isosceles triangle with AC=BC. If AB²=2AC², prove that ABC is a right angled triangle.

Given, ΔABC is an isosceles triangle, ∴ AC=BC (1)
Also, given that, (AB)²=2(AC)²
∴ (AB)²=(AC)²+(AC)² (2)
∴ From (1) and (2), (AB)²=(AC)²+(BC)²
Hence, By converse of Pythagoras theorem, ΔABC is an isosceles right-angled triangle. [hence proved]