Judging from the meaning of the question: ∠ APA1= ∠ BPB1= α AP = a1PBP = b1p.

Then ∠PAA 1 =∠PBB 1 =

∠∠PBB 1 =∠ebf ∴∠pae=∠ebf

≈BEF =∠AEP

∴△BEF ∽△AEP

(2) exists for the following reasons:

Easy to get: △BEF ∽△AEP

If you want to make △ BEF △ AEP, you only need to satisfy BE=AE.

∴∠BAE=∠ABE

* BAC = 60 ∴∠bae=

∠∠ABE =β∠BAE =∠ABE

∴ namely α = 2β+60.

(3) connect BD, and cross A 1B 1 at point g,

Go to A 1 h ⊥ AC at point h through point a1.

≈b 1 a 1p =≈a 1pa = 60 ∴a 1b 1∥ac

From the meaning of the question: AP = A 1P ∠ A = 60.

∴△PAA 1 is an equilateral triangle.

∴a 1h=△Abd, BD= in RT.

∴BG=

∴ (0≤x