The problem is this: Let $g_1$ and $g_2$ be metric tensors in a two-dimensional vector space, if the angle between vectors $v$ and $w$ respecting $g_1$ is the equal to the angle respecting $g_2$, prove that exists $\lambda>0$ such that $g_1 = \lambda g_2$.
At this point, I've tried to reach that conclusion from the formulas regarding angle and metric tensors, but without much success. I´ve been told that there is an easy way to do it involving quadratic forms and the spectral theorem, but I am not sure how to relate these metric tensors to quadratic forms and then use the fact regarding the angles to prove that $g_1 = \lambda g_2$. Any ideas?
HINTS: Start by showing that a $g_2$-orthogonal basis $\{v_1,v_2\}$ for $V$ must be a $g_1$-orthogonal basis. If they are $g_2$-orthonormal, suppose that $g_1(v_1,v_1) = \lambda$ and $g_1(v_2,v_2)=\mu$. Show that you must have $\lambda=\mu$ by considering, say, the vectors $v_1+v_2$ and $v_1-v_2$.