證明: (1)$\because \angle ABC = \angle ADC = 90^{\circ}$
$M，$$N$分別是$AC，$$BD$的中點，$\therefore$在$Rt\triangle ABC$中，$BM=\frac{1}{2}AC，$在$Rt\triangle ACD$中，$DM=\frac{1}{2}AC，$$\therefore BM = DM。$又$\because N$是$BD$的中點，$\therefore MN\perp BD。$
(2)$\triangle MBD$是等腰直角三角形。理由：$\because M$是$AC$的中點，$\therefore AM=\frac{1}{2}AC = BM，$$\therefore \angle BAM = \angle ABM，$$\therefore \angle BMC = 2\angle BAM。$同理可得$\angle DMC = 2\angle DAM。$又$\because \angle BAD = 45^{\circ}，$$\therefore \angle BMD = \angle BMC + \angle DMC = 2(\angle BAM + \angle DAM)=2\angle BAD = 90^{\circ}。$又$\because BM = DM，$$\therefore \triangle MBD$是等腰直角三角形。
