Question

等腰三角形ABC中,AB等于AC,在AB、AC上分别取点E、D,使BC、BD、BE相等,且DE等于AE,

Answer 1

解:DE=AE,则∠EDA=∠A,设∠EDA=∠A=X(度),则∠BED=2X.(三角形外角的性质)

BD=BE,则∠BDE=∠BED=2X,∠EBD=180-∠BDE-∠BED=180-4X.

则∠BDC=∠A+∠EBD=X+180-4X=180-3X;

又BC=BD,则∠C=∠BDC=180-3X.

∠A+∠ABC+∠C=180,即X+(180-3X)+(180-3X)=180, X=36.即角A=36度.