Given, ΔABC is an isosceles triangle having AC = BC and $AB^2 = 2AC^2$
In ΔACB
AC = BC
$AB^2 = 2AC^2$
$AB^2 = AC^2 + AC^2$
= $AC^2 + BC^2$ [Since, AC = BC]
Hence, by Pythagoras theorem ΔABC is right angle triangle
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