Question

已知△ABC的外角∩CBD,∩BCE的平分线相交于点F

Answer 1

∵∠A+∠ABC+ACB=180°

∴∠ABC+∠ACB=180°-∠A

∵∠ABC+∠CBD=180°

∠ACB+∠BCE=180°

∴∠ABC+∠ACB+∠CBD+∠BCE=360°

即180°-∠A+∠CBD+∠BCE=360°

∴∠CBD+∠BCE=180°+∠A

∵∠CBD,∠BCE的平分线相交于点F

∴∠CBF+∠BCF=½﹙∠CBD+∠BCE﹚

=½×﹙180°+∠A﹚

∴∠BFC=180°-½×﹙180°+∠A﹚

=90°-½∠A