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

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=12∠BEF,∠PFE=12∠DFE

∴∠PEF+∠PFE=12（∠BEF+∠DFE）=90°

∵∠PEF+∠PFE+∠P=180°

∴∠P=90°．

故答案为90°．