It is known that in parallelograms ABCD, AC⊥BC, AC = BC, a plane rectangular coordinate system is established with the straight line where BC and AC are located as the coordinate axis, and the straight line of b (-6,0) y = 3x+b passes through point D and intersects with the X axis at point M.

(1) Find the coordinates of m.

(2) Point G is a point on the positive semi-axis of Y axis. When ∠ bgm = 45, find the analytical formula of the straight line where GD is located.

(3) Under the condition of (2), let the straight line GD cross the X axis in n, GM cross the AD in k, and point P starts from point O and moves along the ON-NG at the speed of 1 per second, let the time be t, and let the straight line PA cross the BG in h during the movement. When △GHA is similar to △GKD, find t, and CD = AB can be obtained according to the topic.

It is not difficult to prove that △ACD∽△APM, so MP: AP = 3: 4,

1,

Establish a coordinate system with B as the origin, AB as the X axis and AC as the Y axis.

Then a (-6,0), b (0,0), C(-6,-10), D(- 12,-10), then the CB linear equation is y=4/3? x，

When BP=x, P(-3x/5, -4x/5) can be obtained, then

AP = y =√[(-6+3x/5)& amp； sup2+(4x/5)amp； sup2]=√(x & amp； sup2-36x/5+36)，

To make the point m within the parallelogram, it should be ∠ MPC > 0, ∠ MAD > 0,

When ∠ MPC = 0, AP⊥BC, x= 18/5,

When ∠ mad = 0, p falls at point C, and x= 10,

Therefore, we can get 1, 8/5 < x < 10, 24/5 < y < 8,

2. To make the triangle AMD an isosceles triangle, it should be AD=MD or AM=DM, then

Because m is that A rotates 90 counterclockwise with P as the reference point and then retracts by 3/4, so

xM-xP=(3/4)(yA-yP)，yM-yP=(3/4)(xA-xP)，

So M(-6x/5, -[3x+90]/20),

AD=BC= 10，

MD=√[(-6x/5？ + 12)； sup2+(-[3x+90]/20？ + 10)； sup2],

AM==√[(-6x/5？ +6)& amp； sup2+(-[3x+90]/20？ )& ampsup2],

When AM=MD, the solution is x=5.

When MD=AD, there is no solution.