Solutions:

1.  It appears that the green figure is an ellipse. (Click here to see my animation.)

If this is the case, then DC + DE would have to be constant.  In fact, DC + DE = BC.

Proof:

Since circles C and E are fixed, and therefore have a fixed radius, AB and AC are both constant.  So BC is constant as well since AB + AC = BC

BD + DC = BC

DE = BD (from the isosceles triangle in the construction)
So, DC + DE = BC

2.  Again, it appears that the green figure is an ellipse.  (Click here to see my animation.)

The proof is the same as #1.

3.  It appears that the green figure is a hyperbola. (Click here to see my animation.)

If this is the case, then DC – DE would have to be constant.  In fact, DC – DE = BC.

Proof:

Since circles C and E are fixed, and therefore have a fixed radius, AB and AC are both constant.  So BC is constant as well since AB + AC = BC

DE = BD (from the isosceles triangle in the construction)
DE – DC = BD – DC

BD – DC = –BC

DE – DC = –BC

So, DC – DE = BC