Mathematics - Properties of Triangle Question with Solution | TestHub

To find the possible values of angle C, we start with the given relation between the sides $a$, $b$, and $c$ of $△ABC$:

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

Rearrange the terms to group similar powers:

$$a^4+b^4+c^4-2a^2{c}^2-2b^2{c}^2=0$$

We can rewrite this expression by adding and subtracting $2a^2{b}^2$:

$$a^4+b^4+2a^2{b}^2-2a^2{b}^2+c^4-2a^2{c}^2-2b^2{c}^2=0$$

This allows us to form a perfect square:

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

Now, let's rearrange the terms to form another perfect square involving $c^2$:

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

This is in the form of $(X-Y{)}^2=X^2-2XY+Y^2$, where $X=(a^2+b^2)$ and $Y=c^2$.

So, we can write:

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

Move the $2a^2{b}^2$ term to the right side:

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

Take the square root of both sides:

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

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

Now, we know the Law of Cosines for angle C states:

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

From this, we can express $a^2+b^2-c^2$ as:

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

Substitute this into our derived equation:

$$2ab\cos C=\pm \sqrt{2}ab$$

Since $a$ and $b$ are sides of a triangle, they are positive, so $ab\ne 0$. We can divide both sides by $2ab$:

$$\cos C=\frac{\pm \sqrt{2}ab}{2ab}$$

$$\cos C=\frac{\pm \sqrt{2}}{2}$$

$$\cos C=\pm \frac{1}{\sqrt{2}}$$

This gives us two possible values for $\cos C$:

1. $\cos C=\frac{1}{\sqrt{2}}$

2. $\cos C=-\frac{1}{\sqrt{2}}$

For $\cos C=\frac{1}{\sqrt{2}}$, the angle C in a triangle (where $0<C<\pi$) is:

$$C=\frac{\pi }{4}$$

For $\cos C=-\frac{1}{\sqrt{2}}$, the angle C in a triangle (where $0<C<\pi$) is:

$$C=\frac{3\pi }{4}$$

Both $\frac{\pi }{4}$ and $\frac{3\pi }{4}$ are valid angles for a triangle.

Thus, the possible values of angle C are $\frac{\pi }{4}$ and $\frac{3\pi }{4}$.

The final answer is $\boxed{A,D}$.