Question

如图,已知PA切圆O于A,割线PCD交圆O于C、D,弦AB⊥PC交PD于E,且AE=EC,角APD=30°,若圆O的半径

Answer 1

连结OA,交PD于N,并延长交圆于F,连结BF,∵PA是圆的切线,∴OA⊥AP,〈AEP=90°,（已知）〈APE+〈ANP=90°,〈ANE+〈NAE=90°,∴〈NAE=〈APE=30°,∵AF是直径,∴AF=2R=2√3,∵〈ABF=90°,（半圆上圆周角是直角）,∴BF=AF/2=1,∴AB=√3BF=√3.AE=CE,〈AEC=90°,∴△AEC是等腰RT△,〈EAC=45°,〈ECA=45°,