Triangle $ABC$ has an inscribed circle tangent to sides $AB$, $AC$ and $BC$ at points $C_1$, $B_1$ and $A_1 $ respectively. Let $K$ be a point on the circle diametrically opposite to point $C_1$, $D$ be the intersection point of lines $B_1C_1$ and $A_1K$. Prove that $CD = CB_1$.