As shown in the figure, oabc is a rectangular piece of paper, where OA = 8 and OC = 4. Through folding, point C coincides with point a, and the crease is ef (1). Find out the length of OE (2) And prove (3) whether there is a moving point P in the straight line where EF is located, so that the value of | pb-pc | is the maximum. If not, please explain the reason; if so, calculate the coordinates of point P

As shown in the figure, oabc is a rectangular piece of paper, where OA = 8 and OC = 4. Through folding, point C coincides with point a, and the crease is ef (1). Find out the length of OE (2) And prove (3) whether there is a moving point P in the straight line where EF is located, so that the value of | pb-pc | is the maximum. If not, please explain the reason; if so, calculate the coordinates of point P

① OE = √ (4 & # 178; + 5 & # 178;) = √ 41 (see Table 2 below for the length of CE)
② Aecf is a diamond
According to the Title Meaning: ab & # 178; + EB & # 178; = AE & # 178; = (bc-eb) &# 178;
That is: 4 & # 178; + EB & # 178; = (8-eb) &# 178;
EB=3,CE=5
Similarly, of-3, AF = 5
AE coincides with CE, AF coincides with CF
CF=AF=AE=CE=5
So aecf is diamond
③ When point P coincides with point E, | pb-pa | has a maximum, and the coordinates of point P are (5,4)