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20 1 1 detailed explanation of 25 questions in Shaanxi mathematics examination questions ...
Solution: (1) isosceles.

(2) As shown in Figure ①, connect BE, draw the middle vertical line where BE intersects BC and F, connect EF, and △BEF is the crease triangle of rectangular ABCD.

The bisector of the crease is vertical, AB=AE=2,

Point a is perpendicular to BE, that is, the crease passes through point a.

∴ quadrilateral ABFE is a square.

∴BF=AB=2,

∴F(2,0).

(3) Rectangular ABCD has the largest crease triangle BEF with an area of 4,

The reasons are as follows: ① When F is near BC, as shown in Figure ②.

S△BEF≤ 1? 2? S rectangular ABCD, that is, when F and C overlap, the maximum area is 4.

② When f is on the edge CD, as shown in Figure ③,

F is FH∨BC, AB is at H point, and BE is at K point.

∫S△EKF = 1? 2? KF? AH≤ 1? 2? HF? AH= 1? 2? Rectangular AHFD,

S△BKF= 1? 2? KF? BH≤ 1? 2? HF? BH= 1? 2? S rectangular BCFH,

∴S△BEF≤ 1? 2? S rectangle ABCD = 4.

That is, when f is the midpoint of CD, the maximum area of △BEF is 4.

Let's find the coordinates of point e when the area is the largest.

① When F coincides with point C, as shown in Figure ④.

According to folding, CE=CB=4,

At Rt△CDE, ED=? CE2-CD2? =? 42-22? =2? 3? .

∴AE=4-2? 3? .

∴E(4-2? 3? ,2).

② When f is at the midpoint of the side DC, point E coincides with point A, as shown in Figure ⑤.

At this time, e (0 0,2).

To sum up, when the maximum crease area △BEF is 4, the coordinates of point E are E (0,2) or E(4-2? 3? ,2)