Question

AB=AC,AD=DE=EB,DB=BC,求三角形ABC的内角

Answer 1

∵BE=DE,设∠EBD=∠EDB=α,∴∠AED=∠EBD+∠EDB=2α,∵AD=DE,∴∠A=∠AED=2α,∴∠BDC=∠ABD+∠A=3α,∵BD=BC,∴∠C=∠BDC=3≥,∵AB=AC,∴∠ABC=∠C=3α,又∠A+∠B+∠C=180°,∴8α=180°,

α=22.5°,∴ΔAB的三个角分别为：45°,67.5°、67.5°.