14. Let $A B C D$ be a convex quadrilateral, $F$ a point on the segment $C D$, and $E$ the intersection point of $A C$ with $B F$. It is known that $A B = F C$, $A E = 14$, $B E = 10 \sqrt{2}$, $\widehat{B A C} = \widehat{B F D}$, and $\widehat{B E A} = 45^{\circ}$. What is the length of the segment $E F$?
14. The answer is 6. Extend the segment $B F$ from the side of $F$, and on this extension, consider the point $K$ such that $F K=14$. The angle $\widehat{C F K}$ is equal to the angle $\widehat{D F B}$, as they are vertically opposite angles by construction, so the triangles $A B E$ and $F C K$ are similar and, in particular, congruent, as they have two pairs of equal sides ($A B=C F$ and $A E=F K=14$) and the included angle equal ($\widehat{B A E}=\widehat{D F B}=\widehat{C F K}$). Since the two triangles are congruent, we have that $B E=C K=10 \sqrt{2}$, because they are corresponding sides, and similarly $\widehat{A E B}=\widehat{F K C}=45^{\circ}$ because they are corresponding angles. We also have $\widehat{K E C}=\widehat{A E B}=45^{\circ}$ because they are vertically opposite angles, so the triangle $E C K$ is isosceles with two angles of $45^{\circ}$, and therefore it is particularly a right triangle at $C$. We can then calculate the length of its hypotenuse: $E K=K C \cdot \sqrt{2}=10 \sqrt{2} \cdot \sqrt{2}=20$. By difference, we then obtain $E F=E K-F K=20-14=6$.