Question

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

Answer 1

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Squaring on both sides, we get

${\cos }^{2}C=\frac{{(a^2+b^2-c^2)}^2}{{\left(2ab\right)}^2}=\frac{a^4+b^4+c^4+2a^2{b}^2-2b^2{c}^2-2c^2{a}^2}{4a^2{b}^2}=\frac{2c^2(a^2+b^2)+2a^2{b}^2-2b^2{c}^2-2c^2{a}^2}{4a^2{b}^2}(\because Given{a}^4+b^4+c^4=2c^2(a^2+b^2))=\frac{2c^2{a}^2+2c^2{b}^2+2a^2{b}^2-2b^2{c}^2-2c^2{a}^2}{4a^2{b}^2}=\frac{1}{2}\Rightarrow \cos C=\frac{1}{\sqrt{2}}or\frac{-1}{\sqrt{2}}\Rightarrow C={45}^{\circ }or{135}^{\circ }$