在三角形ABC中;AB=AC,AD,AE分别是角BAC和三角形ABC的外角角BAF平分线BE垂直AE求DA垂直AE和AB与DE是否相等

证明:∠EAB=(1/2)∠BAF; ∠BAD=(1/2)∠BAC.则:∠EAB+∠BAD=(1/2)*(∠BAF+∠BAC)=90度;,即∠EAD=90度,所以,DA⊥AE; 又AB=AC,AD平分∠BAC,则:∠ADB=90度; 又∠BEA=90度,故四边形ADBE为矩形,得AB=DE.(矩形对角线相等)...