Answer:
In \[\Delta \,ABC,\text{ }AB=BC\][ \[\because \] triangle is isosceles] ...(i) In \[\Delta \,ABC\]by pythagoras theorem, \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] \[A{{C}^{2}}=A{{B}^{2}}+A{{B}^{2}}\] [From (i)] \[A{{C}^{2}}=2A{{B}^{2}}\] Hence Proved.
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