Question

如图,三角形ABC中,角ABC=45度,CE垂足于AD于E,BD=DE=1/2DC,求证EA=EC

Answer 1

令∠BAD=α,∠CAD=β,则∠BAC=α+β

tanα=BD/AD=2/6=1/3

tanβ=CD/AD=3/6=1/2

tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)

=(1/2+1/3)/(1-1/2*1/3)=1

又0