Question

如图,已知角abd和角bdc的角平分线交于点e,be交cd与点f,其角1=角3

Answer 1

∠1=∠3,∵∠ABD被BF平分,∴∠1=∠ABF,∴∠ABF=∠3,内错角相等,∴AB∥ CD

∵∠1=∠3,∴BD=DF,又∵DE平分∠BDF,∴DE⊥BF,∴∠3+∠CDE=90°,又∵∠CDE=∠BDE,∴∠2+∠3=90°