(1)∵ |x+y-6|+(x-y)²=0
∴ |x+y-6|=0,  (x-y)²=0
x+y-6=0,  x-y=0
解得：x=3,  y=3
∴點(diǎn)B的坐標(biāo)是(3,3)
 
(2)如圖示：

過B作BE⊥Y軸,垂足為E,過B作BF⊥X軸,垂足為F,
則∠EBF＝90°＝∠ABF+∠ABE
∵BC⊥AB
∴∠ABC＝90°＝∠CBE+∠ABE
∴∠CBE＝∠ABF(同角的余角相等)
由(1)知BE＝BF＝3
又∵∠CEB=∠AFB=90°
∴△CEB≌△AFB(ASA)
∴CB＝AB(全等三角形的對(duì)應(yīng)邊相等)
即：AB＝BC