question_answer

In an isosceles triangle \[ABC,\] if \[\text{AC}=\text{BC}\] and \[\text{A}{{\text{B}}^{\text{2}}}=\text{2A}{{\text{C}}^{\text{2}}}\]then \[\angle C\] is equal to:

A) \[30{}^\circ \]

B) \[45{}^\circ \]

C) \[90{}^\circ \]

D) \[60{}^\circ \]

Correct Answer: C

Solution :

[c] We have, \[AC=BC\]

and     \[A{{B}^{2}}=2A{{C}^{2}}=A{{C}^{2}}+A{{C}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}\] \[[AC=BC]\]

\[\therefore \]By converse of Pythagoras theorem. \[\angle C=90{}^\circ \]