Proof: ∫∠ CAD = (1/2) ∠ bad = 45; ∠EOF=45。
∴∠CAD=∠EOF,AD∥X axis; CD is perpendicular to AD. So CD is perpendicular to the x axis.
Let EG be perpendicular to the X axis at G and ∠ EOF = 45, then EG=OG=(√2/2)OE=2, that is, point E is (2,2).
And tan∠EFO=EG/FG= 1/2, then FG=2EG=4, OF=6. Therefore, point f is (6,0).
(2) When point C and point E coincide, the moving distance of point C is: 2 √ 2-AC/2 = 2 √ 2-2/2 = (3/2) √ 2.
So the value of t is: (3/2) √ 2 √ 2 = 3/2 (seconds).
(3) When 0 seconds ≤t≤ 1/2 seconds: s=( 1/2)t? +( 1/2)t+ 1/8；
When 1/2 seconds
When 3/2 seconds
(4) 12 seconds is just right: point N and point E coincide, and point P and point C coincide.
Let CD and NF intersect at K, and NH is perpendicular to CK at H.
Then NH = CH = 1/2, ∠ KNH = ∠ Efo, Tan ∠ KNH = Tan ∠ Efo = 1/2, that is, KH/NH = 1/2, KH = (.
∴CK = ch+KH = 1/2+ 1/4 = 3/4。 When point N is translated from point E to point F, the value of CK is always equal to 3/4.
Therefore, the time from C to K is 3/4 seconds, when t=2+3/4= 1 1/4 (seconds);
② When the AD point coincides with the X axis, the distance from the N point to the X axis is 1/2, NF = √ 5/2, EN = EF-NF = 2 √ 5-√ 5/2 = (3/2) √ 5.
That is, it took (3/2) √ 5 √ 5 = 3/2 seconds for n to go from E to this position, so t=3/2+2=7/4 (seconds).
Therefore, when point P is inside ⊿OEF, the range of T is: 7/4 second.