Giải thích các bước giải:

a.Vì $ABCD$ là hình bình hành $\to AB//CD,AD//BC$

$\to \widehat{DAF}=\widehat{EBF}, \widehat{AFD}=\widehat{BFE}\to \Delta ADF\sim\Delta BEF(g.g)$

       $\widehat{DCE}=\widehat{EBF}, \widehat{DEC}=\widehat{BEF}\to \Delta BEF\sim\Delta CED(g.g)$
$\to \Delta ADF\sim\Delta CED$

b.Ta có; $BE=BC-CE=2$

$\to \dfrac{FB}{FA}=\dfrac{BE}{AD}=\dfrac13$

$\to \dfrac{FB}{FA-FB}=\dfrac1{3-1}$

$\to \dfrac{FB}{AB}=\dfrac12$

$\to FB=\dfrac12AB=\dfrac{15}2$

$\to AF=AB+BF=\dfrac{45}2$

c.Vì $AB//CD$

$\to \dfrac{CE}{CB}=\dfrac{DE}{DF}$

$\to \dfrac{CE}{DA}=\dfrac{DE}{DF}$

$\to EC\cdot DF=AD\cdot DE$

Vì $\Delta CDE\sim\Delta AFD$

$\to \dfrac{CD}{AF}=\dfrac{CE}{AD}$

$\to CD\cdot DA=AF\cdot EC$