解:(1)延长DM,CB交于点E.(如图3) ∵梯形ABCD中,AD∥BC, ∴∠ADM=∠BEM. ∵点M是AB边的中点, ∴AM=BM 在△ADM与△BEM中, ∠ADM=∠BEM, ∠AMD=∠BME, AM=BM, ∴△ADM≌△BEM, ∴AD=BE=a,DM=EM, ∴CE=CB+BE=b+a ∵CD= , ∴CE=CD ∴CM⊥DM; (2)分别作MN⊥DC,DF⊥BC,垂足分别为点N,F(如图4) ∵CE=CD,DM=EM, ∴CM平分∠ECD. ∵∠ABC= 90°,即MB⊥BC, ∴MN=MB ∵AD∥BC,∠ABC=90°, ∴∠A=90° ∵∠DFB=90°, ∴四边形ABFD为矩形. ∴BF= AD=a,AB= DF ∴FC= BC-BF =b-a ∵Rt△DFC中,∠DFC=90°, ∴ = =4ab ∴ DF=2![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031132223-54635.png) ∴MN=MB= AB= DF=![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031132224-28417.png) 即点M到CD边的距离为![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031132224-95500.png) | ![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031132225-64961.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031132225-84134.png) |