Spin connection equations of motion from Einstein-Palatini action

I am working through "Supergravity" from Freedman and Van Proeyen. In exercise 8.11 one is tasked to vary the Einstein-Palatini action
$$ S = frac{1}{2kappa^2}int d^Dx e e^mu{}_a e^nu{}_b R_{munu}{}^{ab}(omega) $$
wrt. to the spin connection $omega_{mu ab}$ (here greek indices represent "curved" indices, while latin represent "flat" ones).
Ultimately we are to show that this variation is proportional to $D_{[mu}e^a{}_{nu]}$, where $D$ denotes the covariant derivative wrt. the spin connection $omega$, letting us conclude that the solution of the equations of motion is exactly the torsion free connection.
As a hint, we are reminded that we showed earlier
$$delta R_{munu ab} = 2 D_{[mu}deltaomega_{nu]ab}.$$
It is also clear, since $D_mu e = partial_mu e = partial_mu sqrt{-g} = frac{1}{2}sqrt{-g}g^{alpha beta} partial_mu g_{alphabeta} = frac{1}{4}sqrt{-g}partial_mu(g^{alpha beta} g_{alphabeta})= 0$, that we can integrate by parts without worrying about $e$.
Edit 1: As was pointed out in the comments, this is actually wrong. However, below I corrected the calculation so that it does not use this argument.

However I only managed the following (setting $kappa=1$)
$$ delta S = frac{1}{2}int d^Dx e e^mu{}_{a}e^nu{}_b delta R_{munu}{}^{ab} = int d^Dx e e^mu{}_{a}e^nu{}_bD_{[mu}deltaomega_{nu]}{}^{ab} = int d^Dx e e^{[mu}{}_{a}e^{nu]}{}_bD_{mu}deltaomega_{nu}{}^{ab}.$$
We can now rewrite the covariant derivative:
$$ e e^{[mu}{}_{a}e^{nu]}{}_bD_{mu}deltaomega_{nu}{}^{ab} = ee^{[mu}{}_{a}e^{nu]}{}_b(partial_mu deltaomega_{nu}{}^{ab} + omega_{mu}{}^a{}_c deltaomega_{nu}{}^{cb} +omega_{mu}{}^b{}_c deltaomega_{nu}{}^{ac})\
= partial_mu(e e^{[mu}{}_{a}e^{nu]}{}_b deltaomega_{nu}{}^{ab})-partial_mu(e e^{[mu}{}_{a}e^{nu]}{}_b) deltaomega_{nu}{}^{ab} + omega_{mu}{}^c{}_{a}(e e^{[mu}{}_{a}e^{nu]}{}_b) deltaomega_{nu}{}^{ab} + omega_{mu}{}^c{}_b(e e^{[mu}{}_{a}e^{nu]}{}_b)deltaomega_{nu}{}^{ab} \
= partial_mu(e e^{[mu}{}_{a}e^{nu]}{}_b deltaomega_{nu}{}^{ab})- [partial_mu(e e^{[mu}{}_{a}e^{nu]}{}_b) + omega_{mu a}{}^{c}(e e^{[mu}{}_{c}e^{nu]}{}_b) + omega_{mu b}{}^c(e e^{[mu}{}_{a}e^{nu]}{}_c)]deltaomega_{nu}{}^{ab} \
=partial_mu(e e^{[mu}{}_{a}e^{nu]}{}_b deltaomega_{nu}{}^{ab})- D_mu(e e^{[mu}{}_{a}e^{nu]}{}_b)deltaomega_{nu}{}^{ab},$$
where the first term vanishes in the integral and we conclude that
$$D_{mu}(e e^{[mu}{}_{a}e^{nu]}{}_b) =0.$$
however, I do not see how that brings us the proposed solution.
Any hints and help to reach the conclusion would be much appreciated.

Edit 2: I found a source, which skips almost all actual steps of the derivation, but gave one good hint and an intermediate result.
From the definition of the determinant $e$, we can conclude up to factors that:
$$ e e^{[mu}{}_{a}e^{nu]}{}_b sim varepsilon_{abcd}varepsilon^{munurhosigma} e^{c}{}_rho e^d{}_sigma, $$
which then would give
$$D_{mu}(varepsilon_{abcd}varepsilon^{munurhosigma} e^{c}{}_rho e^d{}_sigma) = 0.$$
If now the covariant derivative of the $varepsilon$s would be zero, we would arrive at
$$varepsilon_{abcd}varepsilon^{munurhosigma} D_{mu} (e^{c}{}_rho e^d{}_sigma) = 0,$$
which by anti symmetry immediately gives
$$varepsilon_{abcd}varepsilon^{munurhosigma} D_{mu} (e^{c}{}_rho) e^d{}_sigma = 0,$$
which according to the source, eq. (2.14) is not only the case, but also implies the wanted equations of motion!
However, now I am stuck at proving that actually the covariant derivatives vanish and even if I just assume that, I do not see how then the claimed equations of motion follow.