20. The perimeter of triangle $A B C$, in which $A B<A C$, is 7 times the side length of $B C$. The inscribed circle of the triangle touches $B C$ at $E$, and the diameter $D E$ of the circle is drawn, cutting the median from $A$ to $B C$ at $F$. Find $\frac{D F}{F E}$.
(3 marks)
20. The perimeter of $\triangle A B C$ is 7 times the length of $B C$, and $A B<A C$. $D E$ is the diameter of the incircle of $\triangle A B C$, intersecting $B C$ at $E$. The median from $A$ to $B C$ intersects $D E$ at $F$. Find $\frac{D F}{F E}$.
20. $\frac{5}{7}$
20. Let $M$ be the midpoint of $BC$ and $K$ be the foot of the perpendicular from $A$ to $BC$. Let also $AK=h$, $s$ be the semiperimeter, and $r$ be the radius of the inscribed circle. With standard notations, we have
$$
\begin{array}{l}
CK=b \cos C=\frac{a^{2}+b^{2}-c^{2}}{2a} \\
CE=s-c \\
ME=CE-CM=\frac{b-c}{2} \\
MK=CK-CM=\frac{b^{2}-c^{2}}{2a}
\end{array}
$$

Noting that $\triangle MEF \sim \triangle MKA$, we have $\frac{FE}{AK}=\frac{ME}{MK}=\frac{a}{b+c}=\frac{1}{6}$, i.e., $FE=\frac{1}{6} h$.
On the other hand, we have $\frac{ah}{2}=rs$ since both are equal to the area of $\triangle ABC$.
As a result, $DE=2r=\frac{a}{s} h=\frac{2}{7} h$. So the answer is $\left(\frac{2}{7}-\frac{1}{6}\right): \frac{1}{6}=\frac{5}{7}$.