Question

# If 8R^2 = a^2 + b^2+c^2 ,prove that the triangle is right angle triangle. Here a,b and c are the lengths of side of triangle and R is circum radius.

Right triangles and trigonometry
If $$\displaystyle{8}{R}^{{2}}={a}^{{2}}+{b}^{{2}}+{c}^{{2}}$$ ,prove that the triangle is right angle triangle. Here a,b and c are the lengths of side of triangle and R is circum radius.

2021-01-28
$$\displaystyle{a}={2}{r}{\sin{{A}}}$$
$$\displaystyle{b}={2}{r}{\sin{{B}}}$$
$$\displaystyle{c}={2}{r}{\sin{{C}}}$$
$$\displaystyle{a}^{{2}}+{b}^{{2}}+{c}^{{2}}={4}{R}^{{2}}{\left({{\sin}^{{2}}{A}}+{{\sin}^{{2}}{B}}+{{\sin}^{{2}}{C}}\right)}={8}{R}^{{2}}$$
$$\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{2}}{B}}+{{\sin}^{{2}}{C}}={2}$$
$$\displaystyle{1}-{\cos{{2}}}{A}+{1}-{\cos{{2}}}{B}+{1}-{\cos{{2}}}{C}={4}$$
$$\displaystyle{\cos{{2}}}{A}+{\cos{{2}}}{B}+{\cos{{2}}}{C}+{1}={0}$$
$$\displaystyle{\cos{{A}}}=-{\cos{{\left({B}+{C}\right)}}}$$
$$\displaystyle{1}+{\cos{{2}}}{A}={2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}+{C}\right)}}}$$
$$\displaystyle{2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}+{C}\right)}}}+{\cos{{2}}}{B}+{\cos{{2}}}{C}={2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}+{C}\right)}}}+{2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}-{C}\right)}}}$$
$$\displaystyle{2}{\cos{{\left({B}+{C}\right)}}}{\left({\cos{{\left({B}+{C}\right)}}}+{\cos{{\left({B}-{C}\right)}}}\right)}={0}$$
$$\displaystyle{4}{\cos{{A}}}{\cos{{B}}}{\cos{{C}}}={0}$$
Some of the factors must be zero so one angle is right