Gara di Febbraio

Solution:

The answer is $6$. Extend segment $BF$ beyond $F$, and on this extension consider the point $K$, such that $FK = 14$. The angle $\widehat{CFK}$ is equal to the angle $\widehat{DFB}$, since they are vertical angles by construction, so triangles $ABE$ and $FCK$ are similar and, in particular, congruent, since they have two pairs of equal sides ($AB = CF$ and $AE = FK = 14$) and the included angle equal ($\widehat{BAE} = \widehat{DFB} = \widehat{CFK}$). Since the two triangles are congruent, we have that $BE = CK = 10\sqrt{2}$, being corresponding sides, and likewise $\widehat{AEB} = \widehat{FKC} = 45^\circ$ being corresponding angles. We also have $\widehat{KEC} = \widehat{AEB} = 45^\circ$ since they are vertical angles, so triangle $ECK$ is isosceles with two angles of $45^\circ$, and thus in particular it is right-angled at $C$. We can therefore compute the length of its hypotenuse: $EK = KC \cdot \sqrt{2} = 10\sqrt{2} \cdot \sqrt{2} = 20$. By subtraction we then obtain $EF = EK - FK = 20 - 14 = 6$.