解答:证明:
(1)根据同弧所对的圆周角相等,得∠DAN=∠DBC,∠DCN=∠DBA.
又∵∠DAN=∠BAM,∠BCM=∠DCN,
∴∠BAM=∠MBC,∠ABM=∠BCM.
∴△BAM∽△CBM,
∴
=,即BM
2=AM?CM.①
又∠DCM=∠DCN+∠NCM=∠BCM+∠NCM=∠ACB=∠ADB,
∠DAM=∠MAC+∠DAN=∠MAC+∠BAM=∠BAC=∠CDM,
∴△DAM∽△CDM,
则
=,即DM
2=AM?CM.②
由式①、②得BM=DM,
即M为BD的中点.
(2)如图,延长AM交圆于点P,连接CP.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b03533fa828ba61e65d91ca84234970a304e5906?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴∠BCP=∠PAB=∠DAC=∠DBC.
∵PC∥BD,
∴
=.③
又∵∠MCB=∠DCA=∠ABD,∠DBC=∠PCB,
∴∠ABC=∠MCP.
而∠ABC=∠APC,
则∠APC=∠MCP,
有MP=CM.④
由式③、④得
=.