Question

如图,已知直线PA交圆O于,AB两点...

Answer 1

（1）连接OC．∵点C在⊙O上,OA=OC,∴∠OCA=∠OAC．∵CD⊥PA,∴∠CDA=90°,则∠CAD+∠DCA=90°．∵AC平分∠PAE,∴∠DAC=∠CAO．∴∠DCO=∠DCA+∠ACO=∠DCA+∠CAO=∠DCA+∠DAC=90°．又∵点C在⊙O上,OC为⊙O的半径,∴CD为⊙O的切线．（2）过O作OF⊥AB,垂足为F,∴∠OCD=∠CDA=∠OFD=90°,∴OC=FD,OF=CD．∵DC+DA=6,设AD=x,则OF=CD=6-x,∵⊙O的直径为10,∴DF=OC=5,∴AF=5-x,在Rt△AOF中,由勾股定理得AF^2+OF^2=OA^2．即（5-x）^2+（6-x）^2=25,化简得x^2-11x+18=0,解得x=2或x=9．∵CD=6-x＞0,故x=9舍去,∴x=2,∴AD=2,AF=5-2=3,∵OF⊥AB,由垂径定理知,F为AB的中点,∴AB=2AF=6．