Japan Mathematical Olympiad

17

3

Since $\angle ADE=\angle FDC$ and $\angle DEA=\angle DCF$, triangles $DAE$ and $DFC$ are similar, hence $DA:DE=DF:DC$. Additionally we have $\angle ADF=\angle EDC$ and thus triangles $DAF$ and $DEC$ are similar in the same orientation. This means that the angle formed by lines $AF$ and $EC$ is equal to the angle formed by lines $AD$ and $ED$. Therefore, let $G$ denote the intersection point of lines $AF$ and $BC$, and $\angle AGE=\angle ADE$ holds. Consequently, $\angle BAG=180^{\circ }-\angle ABC-\angle AGB=\angle ADC-\angle AGB=\angle AGB$, which implies $BA=BG$; hence we obtain $EG=BG-BE=2$. Furthermore, since $\angle FGE=\angle FDC$, quadrilateral $DFGC$ is cyclic, and by the power of a point theorem, $EF\cdot ED=EG\cdot EC$ holds. Therefore, with $ED=\frac{26}{3}$, we find $FD=ED-EF=\frac{17}{3}$.