Question

在三角形abc中 ab=ac bd平分角abc 交ac于点d若bd=bc，则角a多少度1疲

Answer 1

设∠A=x，则∠ABC=∠C=（180-x）/2,
∵BD=BC ∴∠BDC=（180-x）/2
∴∠DBC=180-∠BDC-∠BCD=x
∵BD平分∠ABC
∴∠ABD=∠CBD=x
∴∠BCD=∠BDC=∠ABD+∠BAD=x+x=2x
∵∠DBC+∠BCD+∠BDC=180
所以x+2x+2X=180
x=36
∴∠A=36°

Answer 2

设∠A=x
∵AB=AC
∴∠ABC=∠C=(180°-x)/2=90°-x/2
∵BD是∠ABC平分线
∴∠ABD=（90°-x/2）/2=45°-x/4
∵BD=BC
∴∠BDC=∠C=90°-x/2
=∠A+∠ABD
=x+45°-x/4
∴90°-x/2=x+45°-x/4
解得
x=36°
∠A=36°