Question

1.如图,已知△ABC中AB=AC,在AB上取一点D,在CA延长线上取一点E,使AE=AD,连接E

1.如图,已知△ABC中AB=AC,在AB上取一点D,在CA延长线上取一点E,使AE=AD,连接ED并延长交BC于点F,求证DF⊥BC.

Answer 1

很简单啊
∵AD=DE,∴∠E=∠ADE=∠BDF
∵∠B=∠C,由三角形内角和定理得∠BFD=∠CFE
∵∠BFD+∠CFE=180°,∴∠BFD=∠CFE=90°,即DF⊥BC

Answer 2

设<B=a,则<BAC=180-2a=><EDA=(180-2a)/2=90-a=><BDF=90-a,三角形BDF中<DFB=180-(<B+<BDF)=180-90=90度=>DF垂直BC