Question

如图，BD平分∠ABC，BE分∠ABC，BE分∠ABC为2：5两部分，∠DBE=21°，求∠ABC的度数

Answer 1

设∠ABC=x，所以∠CBD=∠DBA=X/2,∠ABE=2/7 X.
所以有：∠DBA - ∠ABE=21
x/2 - 2/7 X =21
X=98
所以：∠CBE=21+∠DBA
=21+ X/2
=70

Answer 2

∵BE分∠ABC为2∶5两部分
∴∠ABE：∠EBC=2：5
∴2∠EBC=5∠ABE
设∠ABD=x
∴∠EBC=x+21°，∠ABE=x°-21°
∴2（x+21°）=5(x-21°）
∴x=49°
∴∠ABC=2x=98°

Answer 3

∠DBC=1/2∠ABC ∠CBE=2/5∠ABC ∠DBE=（1/2∠ABC - 2/5∠ABC )=3/10 ∠ABC
∠DBE=21° ∠ABC =21°÷3/10=70°
∴∠ABC =70°