Đáp án:
Theo Talet ta có:
$\begin{array}{l}
+ DE//AC\\
\Rightarrow \dfrac{{AE}}{{AB}} = \dfrac{{CD}}{{BC}}\\
+ DF//AB\\
\Rightarrow \dfrac{{AF}}{{AC}} = \dfrac{{BD}}{{BC}}\\
\Rightarrow \dfrac{{AE}}{{AB}} + \dfrac{{AF}}{{AC}} = \dfrac{{CD}}{{BC}} + \dfrac{{BD}}{{BC}} = \dfrac{{CD + BD}}{{BC}}\\
\Rightarrow \dfrac{{AE}}{{AB}} + \dfrac{{AF}}{{AC}} = \dfrac{{CD + BD}}{{BC}}\\
\Rightarrow \dfrac{{AE}}{{AB}} + \dfrac{{AF}}{{AC}} = \dfrac{{BC}}{{BC}}\\
\Rightarrow \dfrac{{AE}}{{AB}} + \dfrac{{AF}}{{AC}} = 1
\end{array}$
