Question

如圖所示,已知圓O外有一點(diǎn)P,作圓O的切線PM,M為切點(diǎn),過PM的中點(diǎn)N,作割線NAB,交圓于A、B兩點(diǎn),連接PA并延長,交圓O于點(diǎn)C,連接PB交圓O于點(diǎn)D,若MC=BC.

(1)求證:△APM∽△ABP;
(2)求證:四邊形PMCD是平行四邊形.

Answer 1

見解析

解析證明:(1)∵PM是圓O的切線,NAB是圓O的割線,N是PM的中點(diǎn),
∴MN²=PN²="NA·NB," ∴=,
又∵∠PNA=∠BNP, ∴△PNA∽△BNP,
∴∠APN=∠PBN, 即∠APM=∠PBA.
∵M(jìn)C="BC," ∴∠MAC=∠BAC,
∴∠MAP=∠PAB,
∴△APM∽△ABP.
(2)∵∠ACD=∠PBN,
∴∠ACD=∠PBN=∠APN,即∠PCD=∠CPM,
∴PM∥CD,
∵△APM∽△ABP,∴∠PMA=∠BPA,
∵PM是圓O的切線,∴∠PMA=∠MCP,
∴∠PMA=∠BPA=∠MCP,即∠MCP=∠DPC,
∴MC∥PD,
∴四邊形PMCD是平行四邊形.