证明：连接CO并延长,交圆O于点M,连接BM∵CM是直径∴∠CBM=90°∴∠MCB+∠M=90°∵CD相切与圆O于点C∴∠MCD=90°=∠MCB+∠M又∵∠MCD=∠MCB+∠BCD∴∠MCB+∠BCD=∠MCB+∠M∴∠BCD=∠M∵∠M=∠A∴∠BCD=∠A证毕
