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Variant of Mathematics Final Examination in Senior Three.
(1) First prove △BCE∽△BAO, and get the answer according to the equal ratio of the corresponding sides of two triangles;

(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-.