证明:(1)延长AO交圆于E,连接BE.
∵AE是直径∴角ABE=90°∵∠ABE=∠ADC=90° ∠E=∠C∴△ABE∽△ACD
∴AB/AE=AD/AC∵AE=2AO∴AB*AC=2AD*AO
(2)由△ABE∽△ACD得∠BAE=∠CAD
∵AM平分∠ABC∴∠BAM=∠CAM∴∠OAM=∠DAM即AM平分∠OAD
证明:(1)延长AO交圆于E,连接BE.
∵AE是直径∴角ABE=90°∵∠ABE=∠ADC=90° ∠E=∠C∴△ABE∽△ACD
∴AB/AE=AD/AC∵AE=2AO∴AB*AC=2AD*AO
(2)由△ABE∽△ACD得∠BAE=∠CAD
∵AM平分∠ABC∴∠BAM=∠CAM∴∠OAM=∠DAM即AM平分∠OAD