(2)BQ = AP； BQ⊥AP.

Proof: ○ 1∶ef = FP, EF⊥FP, ∴∠ EPF = 45.

∴∠∵ac⊥bc cqp = 45，

∴CQ=CP.

At △BCQ and △ACP,

BC=AC，∠BCQ = 90° =∠ACP，CQ=CP，

∴△BCQ≌△ACP.

∴BQ=AP.

○2 As shown in figure 18-4, BQ is extended to AP at point M. 。

∵△BCQ≌△ACP,∴∠CBQ=∠CAP.

∠∠CBQ+∠CQB = 90，∠CQB=∠AQM，

∴∠CAM+∠AQM=90，

∴∠ QMA = 90, that is, BQ⊥AP.

(3) established.

Proof: ○ 1 as shown in figure 18-5,

∫∠epf = 45 ,∴∠cpq=45，

∴∠∵ac⊥bc cqp = 45，

∴CQ=CP.

At △BCQ and △ACP,

BC=AC，∠BCQ = 90° =∠ACP，CQ=CP，

∴△BCQ≌△ACP.

∴BQ=AP.

○2 As shown in figure 18-5, extend QB to point N of AP.

∵△BCQ≌△ACP,∴∠CQB=∠APC.

∠∠CBQ+∠CQB = 90，∠PBN=∠CBQ，

∴∠APC+∠PBN=90，

∴∠ QNA = 90, that is, BQ⊥AP.

Note: This is the 24th question in the 2008 senior high school entrance examination in Hebei Province. By observing, measuring, guessing and proving the conclusion, students have experienced the whole process of mathematical discovery and realized the importance of rational reasoning and the necessity of proof.