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.