假设等腰三角形ABC,AB AC分别是两腰,D为AB上中点,E为AC上中点,CD为腰AB中线,BE为腰AC中线,F为DC BE 交点,求证：FB=FC则AB=AC,∠ABC=∠ACB,（先证△DBC≌△ECB）∵AB=AC,D E为AB AC 上中点∴BD=CE又∵∠ABC=∠ACB,BC=...