CD.------------------------------------------------------------------------2分Р(2)证明:∵AB∥CD,∴∠BEF+∠DFE=180°,---------------------------3分Р∵PE、PF分别是∠BEF与∠EFD的角平分线,Р∴∠PEF=∠BEF,∠PFE=∠DFE,Р∴∠PEF+∠PFE=(∠BEF+∠DFE)=90°,РAРBРCРDРMРNРFРPРGРHРQРKР第25题图(3)РEР1Р2Р3Р4Р5Р∴∠EPF=90°,--------------------------------------------------------------------4分Р∵GH⊥EG,∴∠EGH=90°,Р∴∠EPF=∠EGH,Р∴PF∥GH.-----------------------------------------------------------------------5分Р(3)解:∵PF∥GH,∴∠1=∠3,Р∵∠1=∠2,∴∠2=∠3,-------------------------------------------------------6分Р∵PQ平分∠EPK,∴∠4=∠5+∠3+∠2,----------------------------------7分Р由(2)可知∠4+∠5=90°,Р∴∠5+∠3+∠2+∠5=90°,Р∴2(∠5+∠3)=90°,--------------------------------------------------------------8分Р∴∠5+∠3=45°,即∠HPQ==45°,Р∴∠HPQ的大小没有发生变化.---------------------------------------------9分