Đáp án:
Theo Talet ta có:
$\begin{array}{l}
\Delta ADC:EK//DC\\
\Rightarrow \frac{{AE}}{{AD}} = \frac{{AK}}{{AC}}\\
\Delta ABC:KF//AB\\
\Rightarrow \frac{{AK}}{{AC}} = \frac{{BF}}{{BC}}\\
\Rightarrow \frac{{AE}}{{AD}} = \frac{{BF}}{{BC}}\left( { = \frac{{AK}}{{AC}}} \right)
\end{array}$