∴∠ECD=∠ADE=∠AOD=90

∴∠ADO+∠EDC=90，∠OAD+∠ADO=90

∴∠OAD=∠EDC

∴△AOD∽△DCE

②①f is FH⊥OC, OC is H, AB is N,

Judging from the meaning of the question, AB=OC=7, AO=BC=4, OD=5.

∫△AOD∽△DCE

OD:CE=AO:CD ∴CE=2.5,CD=2

∵ Quadrilateral ADEF is a rectangle, DE=AF, ∠ DAB+∠ BAF = 90.

∫oad+∠dab = 90，∴∠OAD=∠BAF.

∴∠EDC=∠BAF

∴△AFN≌△DEC

∴AN=DC=2, FN=EC=2.5, ∴ FH = 6.5 F. The coordinates of the point are (2,6.5).

From A (0,4), F (2 2,6.5) and B (7 7,4), we can get.

c=4 6.5=4a+2b+c 4=49a+7b+c

Solution: a=-0.25 b= 1.75 c=4.

② Solution: Point F is on the parabola found in ①.

The reason is:

According to ① in (2), the expression of parabola is: y=-0.25X square+1.75X+4.

When D(k, 0), then DC=7-k,

Similarly, by △AOD∽△DCE and △ AFN △ DEC.

Find: f (7-k，K(7-K))，

Substitute x = 7-k and y = 4 K(7-K)

So point F is on the parabola found in ①.