三角形ABC中,AB=AC,AE//BC,D为直线AE上任一点,求证DB+DC>2AB

延长BA至F,使BA=AF,连FD
AD||BC,AB=AC,∠FAD=∠ABC=∠ACB=∠DAC.
AF=AC,△FAD≌△CAD,FD=DC,FD+DB>FB
∴BD+DC>2AB