Question

已知，在三角形ABC中，AB=AC,点D在AC上，且BD=BC=AD.求三角形ABC各角的度数

Answer 1

∵ AB=AC
∴∠C=∠ABC
∵ BD=BC
∴∠C=∠BDC
∵ DA=DB
∴∠A=∠ABD

∠A+∠ABC+∠C=180°
∠A+2∠C=180°
∠A+2∠BDC=180°
∠A+2(180°-∠ADB)=180°
∠A+360°-2∠ADB=180°
∠A+360°-2(180°-∠ABD-∠A)=180°
∠A+360°-2(180°-∠A-∠A)=180°
∠A+360°-360°+4∠A=180°
5∠A=180°
∠A=36°
∠B=∠C=(180-∠A)/2=72°

Answer 2

解：因为AB=AC，所以：∠C=∠ABC
又BD=BC，所以：∠C=∠BDC
则：∠BDC=∠ABC
因为BD=AD,所以：∠A=∠ABD
又∠ABC=∠ABD+∠CBD,∠BDC=∠ABD+∠A
所以：∠CBD=∠A=∠ABD
即：∠ABC=∠C=2∠A
又：∠ABC+∠C+∠A＝180°
所以：5∠A=180°
解得：∠A=36°，∠ABC=∠C=2∠A＝72°