Prove that the Galois Group Is $D_{10}$

I've been working on this problem for a couple days and I haven't heard back from any of my support team (TAs/Prof) at Uni. Not looking for an answer, but help and hints would be greatly appreciated. Thank you!

Problem:

Let $$f \in \mathbb{Q}[x]$$ be an irreducible quintic. Let K be it's splitting field. $$f$$ has Galois group $$D_{10}$$ (Dihedral of order 10). Let $$\mathbb{Q}(\sqrt{d})$$ be the unique subfield of K of degree 2 over $$\mathbb{Q}$$.

Let $$\zeta = e^\frac{2\pi}{5}$$. Show that the Galois group of the extension $$K(\zeta):\mathbb{Q}(\zeta)$$ is also Dihedral of order 10 unless $$d = 5$$.

My thoughts so far:

So, I know that $$\mathbb{Q}(\zeta)$$ is a degree four extension over $$\mathbb{Q}$$. I also know that since $$f$$ has Galois group $$D_{10}$$, K must have degree 10 over $$\mathbb{Q}$$.

I know that the subfield structure for K must mirror the subfield structure for $$D_{10}$$. So, first, $$\mathbb{Q}(\sqrt{d})$$ corresponds to $$C_5$$. Also, given this information about $$d$$, we know that $$f$$ has at least one real root.

Basically, to solve the problem, we only need to show that $$|K(\zeta):\mathbb{Q}(\zeta)| = 10$$, since there is only one option for Galois groups of order 10.

Now, our professor told us to look at the subfields of $$K(\zeta)$$, K and $$\mathbb{Q}(\zeta)$$ and to look at $$K \cap \mathbb{Q}(\zeta)$$. This is where I am stuck. I don't know how to think about this intersection because I have no clue how to find what else is in K given the information we have.

Let us fix notations: $$K/\mathbf Q$$ is Galois with group $$\cong D_{10}$$ and $$\mathbf Q(\sqrt d)$$ is its unique quadratic subfield. Denoting by $$\zeta$$ a primitive $$5$$-th root of unity, you want to show that $$K(\zeta)/\mathbf Q(\zeta)$$ is Galois with group $$\cong D_{10}$$ unless $$d=5$$ (more precisely, unless $$5d^{-1}$$ is a square in $$\mathbf Q$$).
It is well known that the cyclotomic extension $$\mathbf Q(\zeta)/\mathbf Q$$ is cyclic of degree $$4$$, hence admits a unique quadratic subfield, which is here its maximal real subfield $$\mathbf Q(\zeta)^+=\mathbf Q(\zeta+{\zeta}^{-1})=\mathbf Q(\sqrt 5)$$. Because $$gcd (4,10)=2$$, it follows that, unless $$d=5$$, the extensions $$\mathbf Q(\zeta)$$ and $$K$$ are linearly disjoint over $$\mathbf Q$$, hence $$Gal(K/\mathbf Q) \cong Gal(K(\zeta)/\mathbf Q(\zeta))$$, and we are done.
NB: For any odd prime $$p$$, it is known from the computation of the discriminant of $$\mathbf Q(\zeta_p)/\mathbf Q$$ that the unique quadratic subfield of $$\mathbf Q(\zeta_p)$$ is $$\mathbf Q(\sqrt {p^*})$$, where $$p^*={(-1)}^{\frac {p-1}{2}}$$, so your problem (and answer) can be generalized when replacing $$5$$ by $$p$$, and the condition $$d\neq 5$$ by $$d\neq p^*$$.