(2) When the bottoms and heights of △ Abd and △ABP are the same, the areas can be equal, so the straight line DP is parallel to AB.

Because the analytical formula of straight line AB is y=x+ 1, we can set the analytical formula of straight line DP as y=x+b and substitute it into the coordinates of point D.

The analytical expression of the straight line DP is y=x- 1. When it is combined with the inverse proportional function y = 2/X, the coordinate of point P is (2, 1).

If P(n.2/n), 2/n=n- 1, that is, 2/n-n=- 1, then 6/n-3n =-3;

In addition, this problem can also be obtained by making PH perpendicular to the X axis and H through the intersection point P, and the area of triangle ABP is equal to the area of triangle ABO plus trapezoid BOHP minus the area of triangle APH. Let P(X.2/X) list the equations.

(3) Connect DN, because EN = ED and Mn = MD, so EN divides DG vertically.

It is easy to prove that ∠NDO=∠DEM, point N is NF⊥x axis is F.