Question

C是AB上一点,三角形ACD和三角形BCE都是等边三角形,BD交CE于G,AE交CD于F.求证：FG//AB

Answer 1

∠ACD=∠BCE=∠ECG=60°∠ACE=∠DCB=120°

AC=CD,BC=CE,△ACE≌△DCB

∠CDB=∠EAC,

AC=CD,∠ACD=∠DCE=60

△ACD≌△DCG,CF=CG,∠CFG=∠CGF=60°

∠FGC=∠BCE

FG‖AB