In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then ∠C = ?

(d) 90°

Given:
AC = BC

$A B^{2}=2 A C^{2}=A C^{2}+A C^{2}=A C^{2}+B C^{2}$

Applying Pythagoras theorem, we conclude that ∆ABC is right angled at C.

or, $\angle C=90^{\circ}$