Question

三角形ABC中角ACB=90度D为AB中点E是AC延长线一点ED交BC于E使角A=3角E证EF=AB

Answer 1

矫正一点：应是“ED交BC于F”.记∠A=3*∠E=3*θ.DC=0.5*AB,∠DCA=∠A=3*θ,从而∠CDE=2*θ.EF=CE/cos(θ)由正弦定理,CE/sin(2*θ)=DC/sin(θ),所以AB=2*DC=CE*sin(θ)/sin(2*θ)=CE/cos(θ) 所以,EF=AB