( 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