Let $K |\Bbb Q$ be a cyclic extension of even degree.Let $f \in \Bbb Q[X]$ be an irreducible polynomial of degree $[K:\Bbb Q]$ having a root in $K$ .Then show that the unique field $F$ such that $\Bbb Q \subset F \subset K$ with $[F:\Bbb Q]=2$ is given by $F=\Bbb Q(\sqrt{D_f})$ where $D_f ( =\prod_{i<j}{(\alpha_i-\alpha_j)^2}$ and $\alpha_i$ are all the roots of $f$ ) is the discriminant.
Since, $\rm{Gal(K|\Bbb Q)}$ is cyclic of even order there exists a unique subgroup of order 2, so that takes care of uniqueness. Since, $K|\Bbb Q$ is Galois, $\alpha_i \in K, \forall i$ .
So I need to show that
(i) $\sqrt{D_f}\in K$ and (ii) $[\Bbb Q(\sqrt{D_f}):\Bbb Q]=2$
Thanks in advance for help!
First of all, there exists a unique intermediate field $F$ such that $[F : \mathbb Q] = 2$ because ${\rm Gal}(K : \mathbb Q)$ has a unique subgroup of index $2$.
As for your actual questions:
(i) $\sqrt{D_f} = \prod_{i < j}(\alpha_i - \alpha_j)$ is clearly generated by the $\alpha_i$'s over $\mathbb Q$, and each $\alpha_i$ is in $K$, so $\sqrt{D_f}$ is in $K$.
(ii) First of all, the discriminant $D_f$ is in $\mathbb Q$. So $[\mathbb Q(\sqrt{D_f}) : \mathbb Q]$ is at most $2$, being a root of the quadratic polynomial $X^2 - D_f \in \mathbb Q[X]$.
So we just need to rule out the possibility that $[\mathbb Q(\sqrt{D_f} ) : \mathbb Q] = 1$. By the Galois correspondence, this means ruling out the possibility that $\sqrt{D_f} = \prod_{i < j}(\alpha_i - \alpha_j)$ is fixed by all automorphisms in ${\rm Gal}(K : \mathbb Q)$. Let $n := {\rm deg}(f) = [K : \mathbb Q]$. Then ${\rm Gal}(K : \mathbb Q)$ is a cyclic subgroup of order $n$ within $S_n$, the group of permutations on the roots $\{ \alpha_1, \dots, \alpha_n \}$. Moreover, it is a transitive subgroup of $S_n$ (since $K$ is a normal extension over $\mathbb Q$, and the $\alpha_i$'s are roots of a single irreducible polynomial $f \in \mathbb Q[X]$). An automorphism fixes $\sqrt{D_f}$ if and only if it is an even permutation of the roots. So you need to prove that the alternating group $A_n$ does not contain a transitive cyclic subgroup of order $n$. Indeed a transitive cyclic subgroup of order $n$ can only be generated by an element with cycle structure $(123\dots n)$. Since $n$ is even, an element with this cycle structure must be odd, which is a contradiction.