In the triangular prism $ABC - A_{1}B_{1}C_{1}$ , $AA_{1} = 2$ , $A_{1}C \perp$ base $ABC$ , $\angle ACB = 90^{\circ}$ , the distance from $A_{1}$ to plane $BCC_{1}B_{1}$ is 1 .
(1) Prove: $AC = A_{1}C$ ;
(2) If the distance between lines $AA_{1}$ and $BB_{1}$ is 2 , find the sine of the angle between $AB_{1}$ and plane $BCC_{1}B_{1}$ .