2 1.(20 12? Yiyang) as shown in figure 1, in a square ABCD with an area of 3, e and f are two points on the side of BC and CD, AE⊥BF is at point G, and be = 1.

(1) verification: △ Abe △ BCF;

(2) Find the area of the overlapping part of △ABE and △BCF (i.e. △ beg);

(3) Now rotate △ABE to △ AB ′ E ′ counterclockwise around point A (as shown in Figure 2), so that point E falls on point E ′ on the edge of CD. Ask whether the area of the overlapping part of △ABE and △BCF has changed before and after rotation. Please explain the reason.

Answer:

(1) proves that ∵ quadrilateral ABCD is a square,

∴∠ABE=∠BCF=90，AB=BC，

∴∠ABF+∠CBF=90，

∵AE⊥BF，

∴∠ABF+∠BAE=90，

∴∠BAE=∠CBF，

In △ABE and △BCF,

∴△ABE≌△BCF

(2) Solution: ∫ The square area is 3,

∴AB=？

In △BGE and △ Abe,

∠∠GBE =∠BAE，∠EGB=∠EBA=90，

∴△BGE∽△ABE

∴? ,

∫BE = 1，

∴AE2=AB2+BE2=3+ 1=4，

∴S△BGE=？ ×S△ABE=？ =?

(3) Solution: unchanged.

Reason: ∵AB=? ，BE= 1，

∴tan∠BAE=？ =? ，∠BAE=30，

AB' = AD, ∠ AB' E' = ∠ ADE' = 90, AE' Male * * *,

∴rt△abe≌rt△ab′e′≌rt△ade′，

∴∠dae′=∠b′ae′=∠bae=30，

∴ab' and AE are on the same straight line, that is, the intersection of BF and AB' is G,

Let the intersection of BF and AE' be h,

Then ∠ Bao = ∠ Hager = 30, and ∠ agB = ∠ AGH = 90, AG male * * *,

∴△BAG≌△HAG

∴S quadrilateral ghe 'b' = s △ ab 'e'-s △ agh = s △ Abe-s △ abg = s △ bge.

The overlapping area of ∴△ABE's with △ △BCF has not changed before and after rotation.

Thank you for adopting it.