
(1)解:連接OD,作OF⊥CD于點(diǎn)F.
∵CD=6,∴

.
∵

.
∴

;
答;

的值為

;
(2)證明:∵

,
∴由(1)知

,
∵CD∥AB,
∴∠CDO=∠DOE.
∴△DOF∽△OED,
∴∠ODE=∠OFD=90°,
∴OD⊥DE,
∴直線DE與半圓O相切.
分析:(1)連接OD,作OF⊥CD于點(diǎn)F,根據(jù)CD=6利用垂徑定理可得DF,再利用OB+BE即可求出OE.然后即可求出

的值;
(2)由(1)知

,利用CD∥AB,求證△DOF∽△OED,可得∠ODE=∠OFD=90°,即可證明直線DE與半圓O相切.
點(diǎn)評:此題主要考查切線的判定,垂徑定理,相似三角形的判定與性質(zhì)等知識點(diǎn),解答此題的關(guān)鍵是作好輔助線:連接OD,作OF⊥CD于點(diǎn)F.這也是此題的突破點(diǎn),此題有一定的拔高難度,屬于中檔題.