Since $AB$ is the external bisector of $\angle DFE,$ it follows that the reflection $P$ of $D$ across $AB$ is the second intersection of $EF$ with the circle $\odot(AEDB)$ with diameter $\overline{AB}.$ $\angle PKF=\angle DKF=\angle KEF$ $\Longrightarrow$ $PK$ is tangent to $\odot(EFK)$ $\Longrightarrow$ $PK^2=PF \cdot PE$ $\Longrightarrow$ $K$ is always the intersection of $\overline{AF}$ with the circle with center $P$ and radius $\sqrt{PF \cdot PE}.$ $KD^2=KP^2=PF \cdot PE=DF \cdot (DF+FE)=DF^2+DF \cdot FE=$ $=DF^2+PF \cdot FE=DF^2+AF \cdot BF.$