Question

如图,OC⊥OA且交圆O于点B,E为圆O上一点,AE交OC于点D,且CE=CD.求证:CE是圆O的切线

Answer 1

证明:连接OE.

CE=CD,则∠CED=∠CDE;

又∠CDE=∠ADO.故∠AED=∠ADO;

OE=OA,则∠OEA=∠OAD.

OC垂直OA,则∠ADO+∠OAD=90度.

所以,∠AED+∠OEA=90度(等量代换).

得CE是圆O的切线.