(2) It is proved that △EDA∽△BOA can be obtained from the equal ratio of the corresponding edges of similar triangles;
(3) The discussion is divided into three situations: m > 0, m=0 and m < 0. When m=0, it must not be true. When m > 0, it can be divided into two cases: 0 < m < 8 and m > 8, which can be solved according to the definition of trigonometric function. When m < 0, point E coincides with point A, and point E does not coincide with point A. 。
Solution: (1) ∫ A (6,0), B (0 0,8),
∴OA=6,0B=8，
AB= = 10，
∠∠CEB =∠AOB = 90，
∴∠OBA=∠EBC，
∴△BCE∽△BAO，
=, that is =,
ce=﹣m+；
(2)∵m=3，
∴BC=8﹣m=5,CE=﹣ m+ =3，
∴BE=4，
∵ point f falls on the y axis (as shown in Figure 2).
,
∴DE∥BO，
∴△EDA∽△BOA，
= =, that is =,
∴OD=，
The coordinate of point d is (0);
(3) Take the midpoint P of CE and cross P as the PG⊥y axis of G point.
∴CP= CE= ﹣ m
(i) when m > 0,
① When 0 < m < 8, as shown in Figure 3,
∠GCP =∠ Bao,
Cos∠GCP = cos∠ Bao =
∴CG=CP？ cos∠GCP= ( ﹣ )= ﹣ m
∴OG=OC+CG=m+ ﹣ m= m+，
According to the meaning of the question, you must
OG=CP
∴ m+ = ﹣ m，
The solution is m=,
② When m≥8, OG > CP obviously does not have the M value that meets the conditions;
(ii) When m=0, point c coincides with the origin o (Figure 4).
;
(iii) when m < 0,
① When point E coincides with point A, as shown in Figure 5,
It is easy to prove that △COA∽△AOB,
∴ = that is =, and the solution is m = ∴;
(2) When point E does not coincide with point A, as shown in Figure 6,
,
OG=OC﹣CG=﹣m﹣( ﹣ m)═﹣ m﹣，
From the meaning of the question, get
OG=CP
That is m = m,
Solve m = |,
To sum up: m or 0 or-or-.