Question

已知,如图,在四边形ABCD中,AD//BC,E是CD中点,且AE平分 角BAD,求证：AD+BC

Answer 1

延长AE交BC延长线于F,连接 BE,

∵AD∥BC,∴∠EAD=∠F,∠D=∠ECF,

∵DE=CE,∴ΔEAD≌ΔEFC,

∴CF=AD,

∵AE平分∠BAD,∴∠EAD=∠EAB,∴∠EAB=∠F,

∴AB=BF,

∴AB=BC+AD.