Question

If in a triangle $ABC$, $a^4+b^4+c^4=2c^2(a^2+b^2)$, prove that $C={45}^o$ or ${135}^o$.

Answer 1

In $\Delta$ABC

$\Rightarrow a^4+b^4+c^4=2c^2(a^2+b^2)$

$\Rightarrow (a^2+b^2{)}^2-2a^2{b}^2+c^4-2c^2{a}^2-2c^2{b}^2=0$

$\Rightarrow (a^2+b^2)+c^4-2c^2{a}^2-2c^2{b}^2=2a^2{b}^2$

$(a^2+b^2-c^2{)}^2=2a^2{b}^2$

Take square root both side we get

$a^2+b^2-c^2=\sqrt{2}ab$

$\frac{a^2+b^2-c^2}{2ab}=\frac{1}{\sqrt{2}}$

From properties of $\Delta ,\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\cos c=\frac{1}{\sqrt{2}}$

$c={45}^o$ or ${135}^o$.