等腰梯形ABCD中,AD平行BC,角ABC等于60度,AC垂直BD,AB等于4,求ABCD周长

过点A作AE⊥BC于E,过点D作DF⊥BC于F,设AC、BD相交于O∵等腰梯形ABCD∴BO＝CO,AB＝CD,∠ABC＝∠DCB∵AC⊥BD∴∠BOC＝90∴∠ACB＝∠DBC＝45∵AE⊥BC∴AE＝CE∵∠ABC＝60,AB＝4∴BE＝2,AE＝2√3∴CE＝AE＝2√3∴BC＝BE+CE＝2√3+2同理可得：CF＝2∴EF＝CE-CF＝2√3-2∵AD∥BC,AE⊥BC,DF⊥BC∴矩形AEFD∴AD＝EF＝2√3-3∴L梯形＝AB+BC+CD+AD＝4+2√3+2+4+2√3-2＝8+4√3