a) Xét $\Delta AB{\text{D}} $ và $\Delta ACE$, ta có: 

$\widehat {A{\text{D}}B} = \widehat {A{\text{E}}C}( = {90^ \circ })$

$\widehat A$ là góc chung

$ \Rightarrow \Delta AB{\text{D}} \sim \Delta ACE(g - g)$

b) Vì $ \Delta AB{\text{D}} \sim \Delta ACE(g - g)$

$ \Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{A{\text{D}}}}{{AE}} \\ \Rightarrow AE.AB = AC.A{\text{D}} \\ $

Theo chứng minh trên: $ \dfrac{{AB}}{{AC}} = \dfrac{{A{\text{D}}}}{{AE}} \\ \Rightarrow \dfrac{{AB}}{{A{\text{D}}}} = \dfrac{{AC}}{{A{\text{E}}}} \\ $

Xét $\Delta A{\text{ED}} $ và $ \Delta ACB$, ta có:

$\dfrac{{AB}}{{A{\text{D}}}} = \dfrac{{AC}}{{A{\text{E}}}}$

$\widehat A$ là góc chung

$ \Rightarrow \Delta A{\text{ED}} \sim \Delta ACB(c - g - c)$

$\Rightarrow \widehat {A{\text{ED}}} = \widehat {ACB}$

c) Ta có: $ \left\{ {\begin{array}{*{20}{c}} {H{\text{D}} \bot AC} \\ {FN \bot AC} \end{array}} \right. \\ \Rightarrow H{\text{D//FN}} \\ $

Xét $\Delta AFN$, ta có: $HD//FN$, theo định lí Ta-let, ta có: 

$\Rightarrow \dfrac{{A{\text{D}}}}{{DN}} = \dfrac{{AH}}{{HF}}(1)$

Mặt khác $ \left\{ {\begin{array}{*{20}{c}} {HE \bot AB} \\ {FM \bot AB} \end{array}} \right. \\ \Rightarrow HE//FM \\ $

Xét $\Delta AFM$,  có: $HE//FM$, theo đinh lí Ta-let, ta có: 

$\dfrac{{AE}}{{EM}} = \dfrac{{AH}}{{HF}}(2)$

$ (1)(2) \Rightarrow \dfrac{{AE}}{{EM}} = \dfrac{{A{\text{D}}}}{{DN}} \\ \Rightarrow MN//DE \\ $

(Định lí Ta-let đảo)