(1) 证明:∵ 等腰梯形ABCD ∴ AB = CD,AC = BD ∵ BC = CB ∴ △ABC≌△DCB(SSS) ∴ ∠BAC =∠CDB ∵ AB = CD 又 ∵ ∠AOB =∠DOC ∴ △AOB≌△DOC(AAS) (2) 解:①过点A作AG⊥BC
∵ △AOB≌△DOC ∴ AO = DO,BO = CO 又 ∵ ∠AOB =∠BOC = 60 ∴ △AOB、△DOC为等边三角形 ∴ DO =" AD" = 4,CO = BC = 8 ∴ AC =" AO" + CO = 12 ∵ ∠OCB = 60,∴∠CAG = 30 ∴ 在Rt△AGC中,
②连结BF,过D作DH⊥BC,交BC于点H ∴ GH =" AD" = 4,BG =" CH" = 在Rt△ABG中, ∵ △ BOC为等边三角形,F是OC的中点 ∴ BF⊥AC 在Rt△AFB中,∵ E是AB的中点 ∴ |