Slovenija 2008

First, let us find the lengths of the segments $AD$ and $CD$. Write $|AD|=v$ and $|CD|=x$. By Pythagoras's theorem $v^2=|AC{|}^2-x^2=|AB{|}^2-(|BC|-x{)}^2$, so $13^2-x^2=15^2-(14-x{)}^2$ or, equivalently, $13^2=15^2-14^2+2\cdot 14\cdot x$. We see that $x=5$ and $v=12$.

Triangles $EDC$ and $ADB$ are similar because $\angle BAD=\angle DEC$. So, $\frac{|EC|}{|CD|}=\frac{|AB|}{|AD|}$ and $\angle DBA=\angle ECD$. This implies $|EC|=\frac{15}{9}\cdot 5=\frac{25}{3}$ and $\angle FCB=\angle CBF$. The triangle $BFC$ is isosceles with the apex at $B$, so $|CF|=|FB|$.



Let $G$ be the midpoint of $BC$. Then $FG$ is perpendicular to $BC$. The triangle $FBG$ is similar to the triangle $ABD$, so $\frac{|FB|}{|BG|}=\frac{|AB|}{|BD|}$ and $|FB|=\frac{|AB|\cdot |BG|}{|BD|}=\frac{15\cdot 7}{9}=\frac{35}{3}$.

The length of the segment $EF$ is $|EF|=|CF|-|CE|=|FB|-|CE|=\frac{35}{3}-\frac{25}{3}=\frac{10}{3}$.