如图,P是⊙O外一点,PA切⊙O于A,PBC是⊙O的割线,AD⊥PO于D、求证:PB/BD=PC/CD.
问题描述:
如图,P是⊙O外一点,PA切⊙O于A,PBC是⊙O的割线,AD⊥PO于D、求证:
=PB BD
.PC CD
答
连接OA,OC,∵PA是切线,∴∠PAO=∠PDA=90°,又∵∠APD=∠OPA,∴△APD∽OPA,∴PDPA=PAPO,∴PA2=PD•PO,又∵PA是切线,∴PA2=PB•PC∴PA2=PD•PO=PB•PC又∵∠CPD=∠OPB,∴△PCD∽△POB∴PCCD=POOB=POOC又△...