( 1) ? As shown in the figure,

? C is the CM⊥x axis, and m is the vertical foot.

? C is the CN⊥y axis and n is the vertical foot.

Then ∠ ncm = 90.

NCB=∠NCM-∠BCM=90 -∠BCM

∠MCA=∠ACB-∠BCM=90 -∠BCM

So ∠NCB=∠MCA

And CB=CA.

∠CNB=∠CMA=90

∴△CNB≌△CMA

So CN=CM

And the coordinates (x, y) of point c satisfy.

x=CN，y=CM

That is x = y.

∴C is on the straight line y = X.

That is, the straight line OC is y = X.

(2)C is on the straight line y = x.

So we can set the c coordinate to (a, a)

Let AM=b, then A(a+b, 0)

B(0，a-b)

So the coordinates of point D are (a+b, a-b).

As shown in the figure, point E is on CE ⊥ de.bd.

Then E(a, a+b)

ED=xD-xE=(a+b)-a=b

CE=yC-yE=a-(a-b)=b

That is ED=CE.

△ CDE is an isosceles triangle.

So DCE = 45.

And ∠ eco = 45.

So ∠ dco = ∠ DCE+∠ ECO = 45+45 = 90.

(3) think about it first