The short answer is that it is counterintuitive (as the discussion so far has proven), but if you integrate the acceleration component in the East-West direction that you are subjecting the helicopter to in the turn you will find out that indeed you will reach a GS of 125KTS by the time you will turn downwind.
Where is the energy coming from? It is the excess energy over straight and level light that you need to supply in order to make a 45 degree bank turn.
Without resorting to the explicit kinematic calculations (which I will be glad to post if requested) consider the following: in a 45 degree banked turn, you are accelerating the helicopter 90 degrees off the direction of travel at 1g (I am not talking about how many g's you are pulling, just what is the sideways acceleration component).
As you start the turn there is no acceleration component in the E-W direction. As you are crossing the midpoint of the turn, you are subjected to a 1g accelleration in the E-W direction, and finally at the end of the turn you are subjected to no acceleration component in the E-W direction.
If you integrate the E-W acceleration component across the turn, you will find that the resulting change in the E-W velocity component will be equal to 2 x 65 KTS = 130 KTS. This means you will go from a groundspeed of 5 KTS to a groundspeed of 125 KTS in the opposite direction, without requiring any excess energy over the one required for a standard turn.
References
A Note on Steep Turns and Wind
Kinematics - Wikipedia, the free encyclopedia