已知∠AOB是圆心角,∠ACB是圆周角,
求证：∠C=1/2∠AOB.
证明：作直径AD,连接BD,
则∠C=∠D(同弧所对的圆周角相等),
∵∠AOB是ΔOBD的外角,
∴∠AOB=∠D+∠OBD,
∵OB=OD,∴∠OBD=∠D,
∴∠AOB=2∠D=2∠C,
∴∠C=1/2∠AOB.