Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {BCD} = {90^0}\\
\widehat {ABH} = \widehat {BDC}\left( {AB//CD} \right)
\end{array} \right.\\
 \Rightarrow \Delta AHB \sim \Delta BCD\left( {g.g} \right)
\end{array}$

b) Ta có:

$\begin{array}{l}
\Delta AHB \sim \Delta BCD\left( {g.g} \right)\\
 \Rightarrow \dfrac{{AH}}{{BC}} = \dfrac{{HB}}{{CD}}\\
 \Rightarrow \dfrac{{AH}}{{HB}} = \dfrac{{BC}}{{CD}}\left( 1 \right)
\end{array}$

Lại có:

$CE$ là phân giác của tam giác $BCD$$ \Rightarrow \dfrac{{EB}}{{ED}} = \dfrac{{CB}}{{CD}}\left( 2 \right)$

Từ $(1),(2)$$ \Rightarrow AH.ED = EB.HB$

c) Kẻ đường cao $CM$ của tam giác $BCD$ ($M\in CD$)

Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHD} = \widehat {CMB} = {90^0}\\
AD = CB\\
\widehat {ADH} = \widehat {CBM}\left( {AD//BC} \right)
\end{array} \right.\\
 \Rightarrow \Delta ADH = \Delta CBM\left( {ch - gn} \right)\\
 \Rightarrow AH = CM
\end{array}$

Khi đó:

$\begin{array}{l}
{S_{AECH}} = {S_{AHE}} + {S_{CHE}}\\
 = \dfrac{1}{2}AH.HE + \dfrac{1}{2}CM.HE\\
 = \dfrac{1}{2}HE\left( {AH + CM} \right)\\
 = AH.HE\left( {do:AH = CM} \right)(*)
\end{array}$

Ta có:

$\begin{array}{l}
\Delta ABD;\widehat A = {90^0};AH \bot BD = H\\
 \Rightarrow \left\{ \begin{array}{l}
BD = \sqrt {A{B^2} + A{D^2}}  = 10cm\\
AH = \dfrac{{2{S_{ABD}}}}{{BD}} = \dfrac{{AB.AD}}{{BD}} = 4,8cm\\
HD = \sqrt {A{D^2} - A{H^2}}  = 3,6cm
\end{array} \right.
\end{array}$

Lại có:

$\begin{array}{l}
\dfrac{{EB}}{{ED}} = \dfrac{{CB}}{{CD}}\\
 \Rightarrow \dfrac{{BD}}{{ED}} = \dfrac{{CB + CD}}{{CD}} = \dfrac{7}{4}\\
 \Rightarrow ED = BD.\dfrac{4}{7} = \dfrac{{40}}{7}cm\\
 \Rightarrow HE = ED - DH = \dfrac{{74}}{{35}}cm
\end{array}$

Thay $AH = 4,8cm;HE = \dfrac{{74}}{{35}}cm$ vảo $(*)$ ta có:${S_{AECH}} = \dfrac{{1776}}{{175}}\left( {c{m^2}} \right)$

Vậy ${S_{AECH}} = \dfrac{{1776}}{{175}}\left( {c{m^2}} \right)$