Question

如图所示,点D在AC上,点E在AB上,且AB=AC,BC=BD,AD=DE=BE,求∠A的度数

Answer 1

∵AB=AC,BC＝BD,AD＝DE＝EB,

∴∠A=∠AED,∠ABC=∠C=∠BDC,∠EDB=∠EBD

设∠A=x,因∠AED=∠EDB+∠EBD

故∠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=180

得X=45°

∴∠A的度数是45°