已知三角形ABC内接圆O,d在oc的延长线上,角B=30度,角cad=30度,求证

1、证明：延长CO交圆O于E,连接AE、OA∵∠B、∠E所对应圆弧都为劣弧AC,∠B＝30∴∠E＝∠B＝30∵OA＝OE∴∠OAE＝∠E＝30∵直径CE∴∠CAE＝90∴∠OAC＝∠CAE-∠OAE＝60∵∠CAD＝30∴∠OAD＝∠OAC+∠CAD＝90∴AD切圆O于A...