Question

三角形内角和应用题

如图，已知：∠A= 1/2∠ABC = 1/2∠C，BD平分∠ABC，求∠DBC的度数
（列式解答~！）

Answer 1

解：
∵∠A+∠ABC+∠C=180°
又∵∠A= 1/2∠ABC = 1/2∠C
∴1/2∠ABC+∠ABC+∠ABC=180°
∴∠ABC=72°
∵BD平分∠ABC
∴∠ABD=∠CBD=1/2∠ABC=36°
我是老师 谢谢采纳

Answer 2

∵∠A+∠ABC+∠C=180°
又∵∠A= 1/2∠ABC = 1/2∠C
∴1/2∠ABC+∠ABC+∠ABC=180°
∴∠ABC=72°
∵BD平分∠ABC
∴∠ABD=∠CBD=1/2∠ABC=36°

Answer 3

解：
∵∠A+∠ABC+∠C=180°
又∵∠A= 1/2∠ABC = 1/2∠C
∴1/2∠ABC+∠ABC+∠ABC=180°
∴∠ABC=72°
∵BD平分∠ABC
∴∠ABD=∠CBD=1/2∠ABC=36°

Answer 4

∵∠A+∠ABC+∠C=180°
又∵∠A= 1/2∠ABC = 1/2∠C
∴1/2∠ABC+∠ABC+∠ABC=180°
∴∠ABC=72°
∵BD平分∠ABC

Answer 5

解：∵∠A= 1/2∠ABC = 1/2∠C
∴∠ABC=∠C=2∠A
∵∠A+∠ABC +∠C=180°
∴∠A+2∠A+2∠A=180°
∴∠A=36°
∵BD平分∠ABC
∴∠DBC=1/2∠ABC
∴∠DBC=∠A=36°