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,DE
2= PD
2+ PE
2,…………………………………………………5分
∴(x+2)
2=2
2+(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x
2=
[2
2+(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.