Question

ab是圆o的直径,d为ac中点,de切圆o于点d,交bc于点e

Answer 1

⑴证明：连接BD,

∵AB为直径,∴BD⊥AC,

∵D为AC中点,∴BD垂直平分AC,

∴AB=BC,∴∠A=∠C,

连接OD,∵OA=OD,∴∠A=∠ODA,

∴∠ODA=∠C,∴OD∥BC,

∵DE是切线,∴OD⊥DE,

∴BC⊥DE.

⑵过O作OF⊥AD于F,

则AF=1/2AD=1/2CD=2,

∵∠A=30°,∴OA=AF/cos30°=4√3/3.