question_answer

In an isosceles \[\Delta \text{ }ABC\] right angled at B, prove that \[A{{C}^{2}}=2A{{B}^{2}}\].

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.