Pentagon $ABCDE$ is inscribed in a circle, and $\angle BEC=\angle CED=45^\circ$. Let line $AC$ and line $BD$ intersect at point $F$, and suppose that $AB=6$ and $AD=8$. What is $BF$?

1. The inscribed angles $\angle BEC$ and $\angle CED$ are both $45^\circ$, so arcs $BC$ and $CD$ each measure $90^\circ$.
2. Hence $\angle BAC=\angle CAD=45^\circ$, so $\overline{AF}$ bisects $\angle BAD=90^\circ$, with $F$ on $\overline{BD}$.
3. By the Pythagorean theorem in $\triangle ABD$, $BD=\sqrt{6^2+8^2}=10$.
4. By the angle bisector theorem, $\dfrac{BF}{FD}=\dfrac{AB}{AD}=\dfrac68=\dfrac34$, so $BF=\dfrac37\cdot10=\boxed{\dfrac{30}7}$.