I think we all need to remember that we are trying to simplify a concept which is actually quite a lot more complex than the basics we teach and learn at flight school. For example the vector diagram (in its various formats depending which book you read) is only considering one particular point along the span of a MR blade.
That said, IMHO, (not being an ADY expert but tending towards the simple (KISS) explanation if possible), there seems to be a fair bit of confusion in some of the responses so far.
Who cares what happens to the A of A in an IGE vs OGE scenario?
In my view, the opinion offered by pofman is closest to the money. If the weight of the helicopter is the same in both scenarios, then the vertical component of the Total Reaction must remain the same. To achieve this situation OGE the Pitch Angle must be increased (due to the increased "Induced Flow") thereby increasing Rotor Drag. The Rotor Drag obviously has to increase (OGE) due to the change in position of the TR. The vertical component of the TR will be the same as it was in the IGE situation but the Rotor Drag component(ie the component of the TR acting in the plane of rotation) will be increased causing an increase in the power required to hover OGE vs that required IGE.
It's easy to see it that way rather than introducing the Lift Formula and its different elements, including A of A.
What does your engine do for you Grasshopper?
Why it overcomes Rotor Drag, Sir.
Good answer son!