解:(1)取中点,连结, 为的中点,,.································ 1分 又,.·································································· 2分 ,得;··············································· 3分 (2)过D作DP⊥BC,垂足为P,∠DAB=∠ABC=∠BPD=90°, ∴四边形ABPD是矩形. 以线段为直径的圆与以线段为直径的圆外切, ,又,∴DE=BE+AD-AB=x+4-2=x+2……4分 PD=AB=2,PE= x-4,DE2= PD2+ PE2,…………………………………………………5分 ∴(x+2)2=22+(x-4)2,解得:. ∴线段的长为.…………………………………………………………………………6分 (3)由已知,以为顶点的三角形与相似, 又易证得.···································································7分 由此可知,另一对对应角相等有两种情况:①;②. ①当时,,.. ,易得.得;················································8分 ②当时,,. .又,. ,即=,得x2=[22+(x-4)2]. 解得,(舍去).即线段的长为2.······································· 9分 综上所述,所求线段的长为8或2. |