Circle, perpendiculars, middle

[ins-]

Let $P$ is a point outside circle $k$. $PA$ and $PB$ are the tangents from $P$ to $k$. A line through $A$ parallel to $PB$ intersects $k$ at the point $C$. The perpendicular from $C$ to $PB$ intersects $k$ and $PB$ at the points $D$ and $E$ respectively. Line through $E$ perpendicular to $BD$ intersects $AC$ at the point $F$. Prove that $AF$=$CF$.

[chessar]

\begin{eqnarray} AC\bot CD\;\Rightarrow\;AD\text{~--- diameter of a circle }k\;\Rightarrow\;AB\bot BD\;\Rightarrow\;EF\parallel AB\;\Rightarrow\;AF=EB\\ \text{Let }G\in PB\colon AG\bot PB\;\Rightarrow\;ACEG\text{~--- rectangle}\;\Rightarrow\;AC=EG\\ \text{It is rather obvious that:}\quad EB=BG\\ \nonumber\text{(1),(2),(3)}\;\Rightarrow\;AF=FC \end{eqnarray}