(20 13? Yantai) The known point P is a moving point on the hypotenuse AB of the right triangle ABC (which is not coincident with A and B), and it passes through the straight line CP in the directions A and B as a vertical line. Solution: solution: (1)AE∑BF, QE=QF.

The reason is: as shown in figure 1, ∫Q is the midpoint of AB,

∴AQ=BQ，

∵BF⊥CP,AE⊥CP，

∴BF∥AE,∠BFQ=∠AEQ=90，

At △BFQ and △AEQ,

∠BFQ=∠AEQ∠BQF=∠AQEBQ=AQ

∴△BFQ≌△AEQ(AAS)，

∴QE=QF，

So the answer is: AE ∨ BF; QE=QF。

(2)QE=QF，

Proof: As shown in Figure 2, extend the intersection of FQ and AE to D.

∫Q is the midpoint of AB,

∴AQ=BQ，

∵BF⊥CP,AE⊥CP，

∴BF∥AE，

∴∠QAD=∠FBQ，

At △FBQ and △DAQ,

∠FBQ=∠DAQBQ=AQ∠BQF=∠AQD

∴△FBQ≌△DAQ(ASA)，

∴QF=QD，

∵AE⊥CP，

Eq is the center line on the hypotenuse of the right triangle DEF,

∴QE=QF=QD，

That is QE = qf.

(3) The conclusion in (2) still holds,

Proof: As shown in Figure 3,

Extend EQ and FB to d,

∫Q is the midpoint of AB,

∴AQ=BQ，

∵BF⊥CP,AE⊥CP，

∴BF∥AE，

∴∠ 1=∠D，

At △AQE and △BQD,

∠ 1=∠D∠2=∠3AQ=BQ，

∴△AQE≌△BQD(AAS)，

∴QE=QD，

∵BF⊥CP，

∴FQ is the center line of hypotenuse DE,

∴QE=QF.