Question

9如图，直线AB∥CD，直线EF分别交AB`CD于E、F，∠BEF的平分线与∠DEF的平分线相交于点P

说明：∠PEF+∠PFE=90°

Answer 1

解：∵AB∥CD

∴∠BEF+∠DFE=180°

又∵∠BEF的平分线与∠DFE的平分线相交于点P

∴∠PEF=12∠BEF，∠PFE=12∠DFE

∴∠PEF+∠PFE=12（∠BEF+∠DFE）=90°

∵∠PEF+∠PFE+∠P=180°

∴∠P=90°．

故答案为90°．