解：(1)∠AOC与∠BOD相等。理由如下：因为A,O,B三点在同一直线上，所以$ \angle AOC + \angle BOC = 180^\circ $（平角定义）。因为$ \angle BOD $与$ \angle BOC $互补，所以$ \angle BOD + \angle BOC = 180^\circ $（互补定义）。因此$ \angle AOC = \angle BOD $（同角的补角相等）。
(2)①因为$ \angle AOC = 32^\circ ，$且$ \angle AOC $与$ \angle MON $互余，所以$ \angle MON = 90^\circ - \angle AOC = 90^\circ - 32^\circ = 58^\circ 。$②$ \angle AON = \angle DON 。$理由如下：设$ \angle AOC = 2x ，$因为OM平分$ \angle AOC ，$所以$ \angle COM = \frac{1}{2}\angle AOC = x 。$因为$ \angle AOC $与$ \angle MON $互余，所以$ \angle MON = 90^\circ - 2x 。$则$ \angle CON = \angle MON - \angle COM = (90^\circ - 2x) - x = 90^\circ - 3x 。$由
(1)知$ \angle BOD = \angle AOC = 2x ，$因为A,O,B共线，所以$ \angle COD = 180^\circ - \angle AOC - \angle BOD = 180^\circ - 2x - 2x = 180^\circ - 4x 。$所以$ \angle DON = \angle COD - \angle CON = (180^\circ - 4x) - (90^\circ - 3x) = 90^\circ - x 。$又因为$ \angle AON = \angle AOC + \angle CON = 2x + (90^\circ - 3x) = 90^\circ - x ，$所以$ \angle AON = \angle DON 。$
