已知,AB是圆o的直径,CB⊥AB于B,AC交圆O于D,DE切圆O圆D交BC于E,求证：DE2=1/4CD*CA

连接BD∵AB是圆o的直径∴BD⊥AC∵CB⊥AB , ∠C=∠C∴△DCB∽△ABC∴CD:BC=BC:AC∴CD*CA=BC^2 ①∵ED和圆O相切,EB和圆O相切∴ED=EB∵OD⊥DE, BD⊥AC,OD=OB∴∠ODB=∠EDC=∠OBD= ∠C∴DE=EC∴DE=BC/2由① CD*CA=BC^2...