Exercise: At a critical stage in the development of an airplane, a decision must be taken to continue or to abandon the project. The financial viability can be measured by a parameter $\theta\in(0,1)$, the project being profitable if $\theta > 1/2$. Data $x$ provide information about $\theta$. If $\theta < 1/2$, the cost of continuing the project is $1/2 - \theta$, whereas if $\theta > 1/2$ the cost is zero. If $\theta > 1/2$ the cost of abandoning the project is $\theta - 1/2$, whereas if $\theta < 1/2$ it is zero.
Denote the decision to continue the project with $d_1$ and denote the decision to abandon the project with $d_2$.
Show that the optimal Bayesian decision is to continue the project if the posterior mean of $\theta$ is greater than $1/2$.
What I've tried: I think that the loss function $L(\theta,d_i)$ looks like this $$L(\theta, d_1) = \begin{cases}0 &\text{ if }\theta > 1/2\\1/2 - \theta &\text{ if }\theta < 1/2\end{cases},\,\,L(\theta, d_2) = \begin{cases}\theta - 1/2 &\text{ if }\theta > 1/2\\0 &\text{ if } \theta < 1/2\end{cases}$$
I know that the decision rule $d$ is Bayes w.r.t. its prior iff it minimises the posterior risk. The posterior risk of $d$ w.r.t. $f_\Theta$(the prior) is defined as $$r_p(f_\Theta,d|x) = \displaystyle\int L(\theta,d(x))f_{\Theta|X}(\theta, x)d\theta.$$ If the posterior mean of $\theta$ is greater than $1/2$ we have $$\displaystyle\int \theta\, f_{\Theta|X}(\theta|x)\,d\theta > 1/2.$$ I think I somehow have to use this fact to show that $\displaystyle\int L(\theta,d_1(x))f_{\Theta|X}(\theta, x)d\theta < \displaystyle\int L(\theta,d_2(x))f_{\Theta|X}(\theta, x)d\theta$ since that would mean that rule $d_1$ is the Bayes rule. However, I'm not sure how.
Question: How do I solve this exercise? Am I headed in the right direction?
Thanks in advance!