如图,三角形abc中,∠a=80度,bf和cf分别是三角形abc的两个外角∠cbd与∠bce的平分线,且交于点f.
1个回答

1、

∵∠A=80,∠ABC=60

∴∠ACB=180-∠A-∠ABC=180-80-60=40

∴∠CBD=180-∠ABC=180-60=120,BCE=180-∠ACB=180-40=140

∵BF平分∠CBD

∴∠FBC=∠CBD/2=60

∵CF平分∠BCE

∴∠FCB=∠BCE/2=70

∴∠F=180-∠FBC-∠FCB=180-60-70=50

2、

∵∠A+∠ABC+∠ACB=180

∴∠ABC+∠ACB=180-∠A

∵∠CBD=180-∠ABC,BF平分∠CBD

∴∠FBC=∠CBD/2=(180-∠ABC)/2=90-∠ABC/2

∵∠BCE=180-∠ACB,CF平分∠BCE

∴∠FCB=∠BCE/2=(180-∠ACB)/2=90-∠ACB/2

∴∠F=180-(∠FBC+∠FCB)

=180-(90-∠ABC/2+90-∠ACB/2)

=(∠ABC+∠ACB)/2

=(180-∠A)/2

∵∠A=80

∴∠F=(180-80)/2=50