Question

设PA.PB是圆O的两条切线,PCD是一条割线,E是AB与PD的交点,证明：PC*DE=PD*CE

Answer 1

由切线定理得PA=PB

由切割线定理得,PA^2=PC*PD=PA*PB

由弦切角定理,△PCB∽△PBD,故PB/PD=CB/BD,

同理,△PCA∽△PAD,故PA/PD=CA/AD

由△ECA∽△EBD得AC/BD=EC/EB

由△EAD∽△ECB得BC/AD=EB/ED

so,

(PB/PD)*(PA/PD)=(PA*PB)/PD^2=PC/PD

(PB/PD)*(PA/PD)=(CB/BD)*(CA/AD)=(AC/BD)*(BC/AD)=(EC/EB)*(EB/ED)=EC/ED

so,PC/PD=EC/ED

即得PC*DE=PD*CE