The circles $k_1, k_2, k_3$ with radii ($a>c>b$) $a,b,c$ are tangent to line $d$ at $A,B,C$, respectively. $k_1$ is tangent to $k_2$, and $k_2$ is tangent to $k_3$. The tangent line to $k_3$ at $E$ is parallel to $d$, and it meets $k_1$ at $D$. The line perpendicular to $d$ at $A$ meets line $EB$ at $F$. Prove that $AD=AF$.