【答案】：
22
【解析】：
设$\angle AOF = x$，因为$OF$平分$\angle AOE$，所以$\angle EOF = \angle AOF = x$。
$\angle DOE$是直角，即$\angle DOE = 90^\circ$，直线$AB$和$CD$相交于点$O$，所以$\angle AOC$与$\angle BOD$是对顶角，$\angle AOC = \angle BOD$。
$\angle COF = 34^\circ$，$\angle AOC = \angle AOF - \angle COF = x - 34^\circ$。
$\angle AOE + \angle DOE + \angle BOD = 180^\circ$（平角定义），$\angle AOE = 2x$，所以$2x + 90^\circ + \angle BOD = 180^\circ$，即$\angle BOD = 90^\circ - 2x$。
又因为$\angle AOC = \angle BOD$，所以$x - 34^\circ = 90^\circ - 2x$，解得$3x = 124^\circ$，$x = \frac{124^\circ}{3}$（此步计算错误，应为$3x = 124^\circ$错误，正确应为$x - 34^\circ = 180^\circ - 2x - 90^\circ$，即$x - 34^\circ = 90^\circ - 2x$，$3x = 124^\circ$错误，正确应为$3x = 124^\circ$不对，重新分析：
$\angle AOD$是平角为$180^\circ$，$\angle AOD = \angle AOF + \angle FOE + \angle EOD$，$\angle AOF = x$，$\angle FOE = x$，$\angle EOD = 90^\circ$，所以$\angle AOD = x + x + 90^\circ = 2x + 90^\circ$，$\angle AOD = 180^\circ$，所以$2x + 90^\circ = 180^\circ$，$2x = 90^\circ$，$x = 45^\circ$，则$\angle AOC = x - \angle COF = 45^\circ - 34^\circ = 11^\circ$，$\angle BOD = \angle AOC = 11^\circ$（又错了，正确应为：
$\angle COE = \angle COF + \angle FOE = 34^\circ + x$，$\angle DOE = 90^\circ$，$\angle COD$是平角$180^\circ$，所以$\angle COE + \angle DOE = 180^\circ$错误，$\angle COD = \angle COE + \angle EOD = 180^\circ$，$\angle COE = 180^\circ - 90^\circ = 90^\circ$，$\angle COE = \angle COF + \angle FOE = 34^\circ + x = 90^\circ$，所以$x = 56^\circ$，$\angle AOF = 56^\circ$，$\angle AOC = \angle AOF - \angle COF = 56^\circ - 34^\circ = 22^\circ$，$\angle BOD = \angle AOC = 22^\circ$。
22
