(1)证明:如图4,由对折和图形的对称性可知,
![](http://img.shitiku.com.cn/uploads/allimg/20191030/20191030000657-93249.jpg) CD=C′D,∠C=∠C′=90° 在矩形ABCD中,AB=CD,∠A=∠C=90° ∴AB=C’D,∠A=∠C’ 在△ABG和△C’DG中, ∵AB=C’D,∠A=∠C’,∠AGB=∠C’GD ∴△ABG≌△C’DG(AAS) ∴AG=C’G (2)解:如图5,设EM=x,AG=y,则有:
![](http://img.shitiku.com.cn/uploads/allimg/20191030/20191030000657-26239.jpg) C’G=y,DG=8-y, DM= AD="4cm " 在Rt△C’DG中,∠DC’G=90°,C’D=CD=6, ∴![](http://img.shitiku.com.cn/uploads/allimg/20191030/20191030000658-81158.gif) 即:![](http://img.shitiku.com.cn/uploads/allimg/20191030/20191030000658-54972.gif) 解得:![](http://img.shitiku.com.cn/uploads/allimg/20191030/20191030000658-77646.gif) ∴C’G= cm,DG= cm 又∵△DME∽△DC’G ∴ , 即:![](http://img.shitiku.com.cn/uploads/allimg/20191030/20191030000659-31594.gif) 解得: , 即:EM= (cm) ∴所求的EM长为 cm。 |