(1) From the known coordinates of point B (2,0), point C (1, 1) and point D (2,4),

The coordinates of point C are (1, 1), and it is easy to get that the resolution function of straight line OC is y=x,

Therefore, the coordinate of point m is (2,2),

So S△CMD= 1, S- trapezoid ABMC=32.

So s △ cmd: s trapezoid ABMC = 2: 3,

That is, conclusion ① holds.

Let the resolution function of straight CD be y=kx+b,

Then k+b = 12k+b = 4,

The solution is k = 3b =? 2

So the resolution function of straight CD is y = 3x-2.

From the above, it can be seen that the coordinates of point H are (0, -2) and yH=-2.

Because of xC? xD=2，

So xC? xD=-yH，

That is, conclusion ② holds;

(2) The conclusion of (1) still holds.

Reason: When the coordinate of A is (t, 0) (t > 0), the coordinate of point B is (2t, 0), the coordinate of point C is (t, t2), and the coordinate of point D is (2t, 4t2).

The coordinate of point C is (t, t2), and it is easy to get that the resolution function of straight line OC is y=tx.

Therefore, the coordinate of point M is (2t, 2t2).

So S△CMD=t3, S- trapezoid ABMC=32t3. ..

So s △ cmd: s trapezoid ABMC = 2: 3,

That is, conclusion ① holds.

Let the resolution function of straight CD be y=kx+b,

Then tk+b = t22tk+b = 4t2,

K = 3tb =？ 2t2

Therefore, the resolution function of straight CD is y = 3tx-2t2;

It can be seen from the above that the coordinates of point H are (0, -2t2) and yH=-2t2.

Because of xC? xD=2t2，

So xC? xD=-yH，

That is, conclusion ② holds;

(3) From the meaning of the question, when the quadratic resolution function is y = ax2 (a > 0) and the coordinates of point A are (t, 0) (t > 0), the coordinates of point C are (t, at2) and the coordinates of point D are (2t, 4at2).

Let the analytical formula of straight line CD be y=kx+b,

Then: tk+b = at22tk+b = 4at2,

Get k = 3atb =？ 2at2

Therefore, the resolution function of the straight line CD is y=3atx-2at2, the coordinates of the point H are (0, -2at2), and YH =-2at2.

Because of xC? xD=2t2，

So xC? xD=- 1ayH。