∵RE△AB'C' is derived from the rotation of RT△ABC.

∴AC=AC' AB=AB' ∴△ABB' and △CAC' are isosceles triangles.

∠CAB=∠C'AB' (constant rotation angle) ∠CAB+∠BAC'=∠C'AB' +∠BAC'

That is, ∠CAC'=∠BAB' (because it is an isosceles triangle, the other angles are equal)

∴△ABB' similar to△△△ CAC'

∴∠ACE=∠FBE ∠BEF=∠CEA。

So △△ACE is similar to △△FBE.

(2)

If 2α=β. . ,

CA=AC '，∠CAC'=β=2α

∴∠ACC'=( 180 -2α)/2=90 -α

∠ ACC' +∠ BCE = 90。

∴∠ECB=α=∠ABC

That is, BE=EC.

△ACE is similar to △FBE in (1).

So △ACB and △FBC are congruent triangles.

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I wrote all the above myself. Many people wrote this question in a perfunctory way on the internet. It seems that they have a faint smell of pretending not to understand. . .

I happen to be doing 20 12 Xiaogan senior high school entrance examination math simulation volume 9.

Just passing by, the answer is as above, I hope to understand.