Question:
In an isosceles triangle $\mathrm{ABC}$ if $\mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$, then $\angle \mathrm{C}=$
(a) $30^{\circ}$
(b) $45^{\circ}$
(C) $90^{\circ}$
(d) $60^{\circ}$
Solution:
Given: In Isosceles $\triangle \mathrm{ABC}, \mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$.
To find: Measure of angle C
In Isosceles ΔABC,
AC = BC
∠B=∠A (Equal sides have equal angles opposite to them)
$\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}$
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}(\mathrm{AC}=\mathrm{BC})$
$\Rightarrow \triangle \mathrm{ABC}$ is right angle triangle, with $\angle \mathrm{C}=90^{\circ}$
Hence the correct answer is $(c)$.