LBC:0=-a+b,∴a=b，

That is, y = bx+B.

Even OD, ∫△ABD is a right triangle, OD is the center line of hypotenuse AB,

From |AB|=2, ∴OD= 1 (unchanged)

(2)① Because ∠ABC is a public angle,

∴△ABD∽△CBO。

BD/BO=BC/BA，

x/y=√( 1+b？ )/2

y=2x/√( 1+b？ )

② when x = 1, it is given by DB=OB= 1 and BC=BA=2.

AD=OC=√3, that is, b=√3,

LBC:y=√3x+√3。