如图，△ABC中，点D,E分别在AC,AB上，AB=AC,BC=BD,AD=DE=EB.求∠A的度数。

显示全部楼层 · 2011-9-21 17:44:10

∠A=180°-2∠C=180°-2∠BDC.....................1∠BDC=∠A+∠ABD...................................2∠A=2∠ABD..............................................3由2，3得∠BDC=∠A+∠A/2=3∠A/2........................4由1，4得∠A=180°-2*3∠A/2=180°-3∠A所以4∠A=180°∠A=45°...

千问 · 2011-9-21 17:44:10

证明：∵AB=AC,BC＝BD，AD＝DE＝EB，∴∠A=∠AED,∠ABC=∠C=∠BDC, ∠EDB=∠EBD设∠A=x°∵∠AED=∠EDB+∠EBD∴∠EDB=∠EBD=x/2∵∠BDC=∠A+∠ABD=x+x/2=3x/2∴∠BDC=∠C=∠ABC=3x/2∵∠A+∠ABC+∠C=180°∴x+3x/2*2=18...