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I'm studying theory of quadratic forms on Serre's book "A course in Arithmetic". I'm trying to prove Theorem 2, chapter 4, which says that given two orthogonal bases $E$ and $F$ for a quadratic module $(V,Q)$ one can find a chain of orthogonal contiguous bases (each one has an element in common with the following bases) relating $E$ and $F$.
He divides the proof in 3 cases, the last being the most difficult. I'm ok with the proof in general, but there is a step that I can't understand, I guess it's a very simple thing but I'm in trouble.

The case in which I'm interested is the following: given the bases $E=e_1, dots, e_n$ and $F=f_1, dots, f_n$, we suppose that $(e_1.e_1)(f_i.f_i)-(e_1.f_i)^2 =0$ for $i=1,2$, where $(x.y)$ is the scalar product associated with the quadratic form. Then we can demonstrate that there is an element $x in mathbbK$ such that $f_x=f_1+xf_2$ is not isotropic and the plane generated by $e_1,f_x$ is nondegenerate. So I take this $x$ and I have that $mathbbKf_1 oplus mathbbKf_x=mathbbKf_1oplus mathbbKf_2$. Having $f_1$ not isotropic since it is part of an orthogonal bases, I can find $f'$ such that $f_x,f'$ is an orthogonal bases of $mathbbKf_1oplus mathbbKf_2$. Then $f_x,f',f_3, dots, f_n$ is a bases for $V$ and $f_x$ is orthogonal with all the $f_i$ by a simple calculation. But I can't see why $f'$ is orthogonal with all the $f_i$'s. I already apologize because I'm quite sure there is a simple explanation. Any hel will be appreciated. Thanks!