Question

问道初二题：如图,CB、CD分别是钝角△AEC和锐角△ABC中线,且AC=AB,∠ACB=∠ABC.求证CE=2CD.

Answer 1

延长CD至F,使DF=CD,连接AF,

AD=BD,CD=DF,∠ADF=∠BDC,

∴⊿ADF≌⊿BDC,

∴AF=BC,AF∥BC

∴∠CAF+∠ACB=180°,∵ ∠ACB=∠ABC,∠ABC+∠CBE=180°

∴∠CAF=∠CBE

又因为AC=BE,

∴⊿CAF≌⊿CBE

∴CE=CF

∴CE=2CD