Question

如图过圆O为一点P作圆的一条切线PC和一条割线PAB,已知这两条线均在PO的同一侧,CQ垂直PO于Q,求证QC平分∠AQ

Answer 1

证明:连接QB,OB.

∠PCO=∠PQC=90°;∠CPQ=∠OPC.则⊿PCO∽⊿PQC,PC/PQ=PO/PC,PC²=PQ*PO.

PC为切线,则:PC²=PA*PB.

∴PA*PB=PQ*PO,PA/PQ=PO/PB;又∠APQ=∠OPB.

则⊿APQ∽⊿OPB,∠PQA=∠PBO.------------------------------------(1)

同理可证:⊿OQC∽⊿OCP,可得:OC²=OQ*OP.则OB²=OQ*OP,OB/OQ=OP/OB.

又∠BOQ=∠POB,则⊿BOQ∽⊿POB,∠BQO=∠PBO.------------(2)

故:∠PQA=∠BQO.

所以∠AQC=∠BQC(等角的余角相等),即QC平分∠AQB.