Question

1.如图,已知BD,BE分别为△ABC的角平分线和高,∠ABC=∠ACB=2∠A,求∠EBD的度数.

Answer 1

∵∠ABC+∠ACB+∠A＝180,∠ABC=∠ACB=2∠A

∴2∠A+2∠A+∠A＝180

∴∠A＝36

∴∠ABC＝∠ACB＝2∠A＝72

∵BD⊥AC

∴∠CBD+∠ACB＝90

∴∠CBD＝90-∠ACB＝90-72＝18

∵BE平分∠ABC

∴∠CBE＝∠ABC/2＝72/2＝36

∴∠EBD＝∠CBE-∠CBD＝36-18＝18°