∵BE平分∠ABM,∴∠ABE=1/2∠ABM,
∵AC平分∠OAB,∴∠BAC=1/2∠OAB,
∵∠ABM=∠O+∠OAB,
∠ABE=∠C+∠BAC,
∴2(∠C+∠BAC)=∠O+2∠BAC,
∴∠O=1/2∠C=45°为定值.没看懂∵BE平分∠ABM，∴∠ABE=1/2∠ABM，.......已知，∵AC平分∠OAB，∴∠BAC=1/2∠OAB，.......角平分线定义∵∠ABM=∠O+∠OAB，.......三角形的外角等于与它不相邻的两个内角和，∠ABE=∠C+∠BAC，.........同上∴2(∠C+∠BAC)=∠O+2∠BAC，.......替换∴∠O=1/2∠C=45°为定值。∵BE平分∠ABM，∴∠ABM=2∠ABE，....①∵AC平分∠OAB，∴∠OAB=2∠BAC，∵∠ABM=∠O+∠OAB=∠O+2∠BAC，∠ABE=∠C+∠BAC，(代入①得)∴2(∠C+∠BAC)=∠O+2∠BAC，∴∠O=1/2∠C=45°为定值。