问题补充:
如图,在△ABC中,AB=AC,D为AC边上一点,且BD=BC=AD,则∠A等于
答案:
∵BD=BC∴∠BDC=∠C∵AB=AC∴∠ABC=∠C∴∠A=∠DBC
∵AD=BD∴∠A=∠DBA∴∠A=∠DBA=∠DBC=1/2∠ABC=1/2∠C
∵∠A+∠ABC+∠C=5∠A=180°∴∠A=36°
======以下答案可供参考======
供参考答案1:
∵BD=BC∴∠BDC=∠C∵AB=AC∴∠ABC=∠C∴∠A=∠DBC
∵AD=BD∴∠A=∠DBA∴∠A=∠DBA=∠DBC=1/2∠ABC=1/2∠C
∵∠A+∠ABC+∠C=5∠A=180°∴∠A=36°
供参考答案2:
∵BD=BC=AD,
∴△ABD,△BCD为等腰三角形,
设∠A=∠ABD=x,则∠C=∠CDB=2x,
又∵AB=AC可知,
∴△ABC为等腰三角形,
∴∠ABC=∠C=2x,
在△ABC中,∠A+∠ABC+∠C=180°,
即x+2x+2x=180°,
解得x=36°,
即∠A=36°.
故本题答案为:36°.
供参考答案3:
.