∴∠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.