13. In the triangle $A B C$, a circle passes through the point $A$, the midpoint $E$ of $A C$, the midpoint $F$ of $A B$ and is tangent to the side $B C$ at $D$. Suppose
$$
\frac{A B}{A C}+\frac{A C}{A B}=4 .
$$

Determine the size of $\angle E D F$ in degrees.
13. Answer: 120
Solution.

Let $B C=a, C A=b$ and $A B=c$. Let $B D=a_{1}$ and $D C=a_{2}$. Using the power of $B$ with respect to the circle, we have $a_{1}^{2}=c^{2} / 2$. Similarly, $a_{2}^{2}=b^{2} / 2$. Thus $b+c=\sqrt{2}\left(a_{1}+a_{2}\right)=$ $\sqrt{2} a$, or $2 a^{2}=(b+c)^{2}$. Therefore,
$$
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{2 b^{2}+2 c^{2}-(b+c)^{2}}{4 b c}=\frac{1}{4}\left(\frac{b}{c}+\frac{c}{b}-2\right)=\frac{1}{4}(4-2)=\frac{1}{2} .
$$

Therefore, $\angle A=60^{\circ}$. Since $A, F, D, E$ are concyclic, $\angle E D F=120^{\circ}$.