连接OC,过点O作OF⊥AC于F∵CD⊥PA,OF⊥AC∴∠ADC＝∠AFO＝90∵AC平分∠PAE∴∠PAC＝∠OAC∴△ACD∽△AOF∴AF/OF＝AD/CD∵CD＝2AD∴AD/CD＝1/2∴AF/OF＝1/2∴OF＝2AF∵⊙O的直径为10∴AO＝5∵OF⊥AC∴AO²＝AF
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