Lời giải:
Ta có:
$DE//BC\quad (gt)$
$\Rightarrow \mathop{DB}\limits^{\displaystyle\frown} = \mathop{EC}\limits^{\displaystyle\frown}$
$\Rightarrow \widehat{DAB} = \widehat{DCB} = \widehat{EAC} = \widehat{FAC}$
Mặt khác:
$\widehat{DBA} = \widehat{DCA}$ (cùng chắn $\mathop{AD}\limits^{\displaystyle\frown}$)
$\Rightarrow \widehat{DBA} + \widehat{DAB} = \widehat{DCA} + \widehat{DCB}$
$\Rightarrow \widehat{DBA} + \widehat{DAB} = \widehat{ACB}$
$\Rightarrow \widehat{DBA} + \widehat{DAB} = \widehat{FAC} + \widehat{AFC}$
$\Rightarrow \widehat{DBA} = \widehat{AFC}$
Xét $ΔABD$ và $ΔAFC$ có:
$\begin{cases}\widehat{DAB} = \widehat{FAC}\quad (cmt)\\\widehat{DBA} = \widehat{AFC}\quad (cmt)\end{cases}$
Do đó: $ΔABD \sim ΔAFC\, (g.g)$
$\Rightarrow \dfrac{AB}{AF} = \dfrac{AD}{AC}$
$\Rightarrow AD.AF = AB.AC$
