零五网 › 全部参考答案› 通城学典课时作业本答案 › 通城学典课时作业本九年级数学人教版南通专版 › 第11页

第11页

信息发布者：

$2\sqrt{5}-2$

解: （1）在Rt△ABC中，$\angle ACB = 90^{\circ}，$$\angle A = 36^{\circ}，$$\therefore \angle ABC = 54^{\circ}.$ 由旋转的性质，可知$BC = CE，$$\angle ACB = \angle DCE = 90^{\circ}，$$\angle ABC = \angle E = 54^{\circ}，$$\therefore \angle CBE = \angle E = 54^{\circ}.$ $\therefore \angle BCE = 72^{\circ}.$ $\therefore \angle BCD = 18^{\circ}$ （2）设AB与CD的交点为H，连接CM. 由旋转的性质，可知$AC = CD，$$\angle A = \angle D = 36^{\circ}=\angle ACD.$ $\therefore AH = CH，$$DE// AC.$ $\therefore \angle A = \angle DMA = 36^{\circ}.$ $\therefore \angle D = \angle DMA.$ $\therefore DH = HM.$ $\therefore AM = DC = AC.$ $\therefore \angle AMC = \angle ACM = 72^{\circ}.$ $\therefore \angle DCM = 36^{\circ} = \angle D.$ $\therefore DM = CM.$ $\because \angle DCE = 90^{\circ}，$$\therefore \angle DEC = \angle MCE = 54^{\circ}.$ $\therefore ME = MC.$ $\therefore DM = ME.$ $\therefore M$是DE的中点