Why is it the middle ring of anthracene which reacts in a Diels–Alder?
To provide a reason for the observed regioselectivity, it is helpful to draw anthracene's aromatic π-electron system in alternance of single and double bonds.  In this instance, it is more beneficial than "the ring" symbolizing the delocalised electron system, as this helps you to account for the precise number of π-electrons before the reaction (starting materials), during the reaction (the mechanism), and after the reaction (the product).



For the Diels–Alder reaction, you may imagine two different pathways.  I invite you to draw the mechanisms by yourself:


A reaction that involves carbon atoms #1 and #4 (or #5 and #8).  Possible, by mechanism.  There are five double bonds remaining in conjugation, and you count one six-membered ring in the state of "a benzene ring" (the very left one).  In the very right six-membered ring, there is only a single double bond, too.
Alternatively, a Diels–Alder reaction with carbon atoms #9 and #10.  This is more favourable then the former example, because both the very left as well as the very right ring of the product are in the state of "a benzene ring", each of these rings contain 6 π-electrons, the prerequisite of a Hückel aromatic compound.  It is so much favourable to the former, that this is the reaction observed.


It may be helpful to add that benzene, naphthalene and anthracene are of course Hückel-aromatic compounds; with 6, 10 or 14 π-electrons they fit into the rule of $(4n + 2)$.  Yet gradually, as experimentally found, in this group of three, benzene is the most, anthracene the least aromatic compound.