Question

已知：如图，AB ∥ CD，直线EF分别交AB、CD于点E、F，∠BEF的平分线与∠DEF的平分线相交于点P．∠P=__

Answer 1

∵AB ∥ CD

∴∠BEF+∠DFE=180°

又∵∠BEF的平分线与∠DFE的平分线相交于点P

∴∠PEF=

1

2 ∠BEF，∠PFE=

1

2 ∠DFE

∴∠PEF+∠PFE=

1

2 （∠BEF+∠DFE）=90°

∵∠PEF+∠PFE+∠P=180°

∴∠P=90°．

故答案为90°．