It is quite known fact that the determinant of arbitrary symmetric matrix is an irreducible polynomial in algebra $\mathbb C [x_{ij}, 1\leq i,j\leq n]$ ($x_{ij}=x_{ji}$) (see this: atlas.mat.ub.es/personals/sombra/papers/cayley/cayley.ps ).

Is there any geometric proof of this statement like the proof of irreducibility of determinant (from the biduality theorem in Gelfand-Zelevinsky-Kapranov book)?

Upd: I have found an algebraic proof of this fact (But I still need geometric). Since $\det (A A^{T})=\det(A)^2$ our polynomial (if not irreducible) is a square of irreducible. Since $\det diag(a_1,\ldots ,a_n)=a_1\cdot\ldots\cdot a_n$ our polynomial cannot be a square of any polynomial.