It is proved that EF divides BC vertically,

∴BF=FC,BE=EC，

∴∠ 1=∠2,

∫∠ACB = 90 degrees,

∴∠ 1+∠4=90 ,∠3+∠2=90 ,

∴∠3=∠4,

∴EC=AE，

∴BE=AE，

CF = AE，

∴BE=EC=CF=BF，

∴ Quadrilateral BECF is a diamond.

Solution: (2) When ∠ A = 45, the diamond BECF is a square.

Proof: ∫∠A = 45, ∠ ACB = 90,

∴∠ 1=45 ,

∴∠EBF=2∠A=90，

Is a diamond a square?

5.( 1) If you want to buy x cars, then you need to buy (10-x) vans.

7x+4( 10-x)≤55，

The solution is x≤5.

And because x≥3, then x = 3, 4 or 5.

Therefore, there are three car purchase schemes:

Option 1: 3 cars and 7 vans;

Option 2: 4 cars and 6 vans;

Option 3: 5 cars and 5 vans.

(2) The daily rent for Option 1 is:

3× 200+7×110 =1370 (yuan);

The daily rent for option 2 is:

4× 200+6×110 =1460 (yuan);

The daily rent for Option 3 is:

5× 200+5×110 =1550 (yuan).

Therefore, in order to ensure that the daily rent is not lower than 1500 yuan, the third option should be selected.

(1) proves that ∵ABCD is a square,

∴AD=BC,∠ADC=∠BCD=90，

Triangle CDE is an equilateral triangle,

∴CE=CD,∠EDC=∠ECD=60，

∴∠ADE=∠ECB，

∴△ade≌△bce；

(2) Solution: ∫△CDE is an equilateral triangle,

∴CE=CD=BC，

△ CBE is an isosceles triangle,

Vertex angle < ECB = 90 < 60 = 30,

∴∠EBC=( 180 ﹣30 )=75，

∫ AD ∨ BC,

∴∠AFB=∠EBC=75。