Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {ABH} = \widehat {BDC}\left( {AB//CD} \right)\\
\widehat {AHB} = \widehat {BCD} = {90^0}
\end{array} \right.\end{array}$
$\Rightarrow \Delta ABH \sim \Delta BDC\left( {g.g} \right)$
b) Ta có:
$\Delta ABH \sim \Delta BDC\left( {g.g} \right)\\
\Rightarrow \dfrac{{AH}}{{BC}} = \dfrac{{BH}}{{DC}}\\
\Rightarrow \dfrac{{AH}}{{BH}} = \dfrac{{BC}}{{DC}}\left( 1 \right)$
Và:
$CE$ là tia phân giác của $\widehat {BCD}$
$ \Rightarrow \dfrac{{CB}}{{CD}} = \dfrac{{EB}}{{ED}}\left( 2 \right)$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow \dfrac{{AH}}{{BH}} = \dfrac{{EB}}{{ED}}$
$ \Rightarrow AH.ED = BH.EB$
c) Kẻ $CK\bot BD=K$
Ta có:
$\begin{array}{l}
{S_{AECH}} = {S_{AHE}} + {S_{CHE}}\\
= \dfrac{1}{2}HE.AH + \dfrac{1}{2}HE.CK\left( 3 \right)
\end{array}$
Mà:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {CKD} = {90^0}\\
AB = CD\\
\widehat {ABH} = \widehat {CDK}\left( {AB//CD} \right)
\end{array} \right.\\
\Rightarrow \Delta AHB = \Delta CKD\left( {ch - gn} \right)\\
\Rightarrow AH = CK\left( 4 \right)
\end{array}$
Từ $\left( 3 \right),\left( 4 \right) \Rightarrow {S_{AECH}} = AH.HE\left( * \right)$
Lại có:
$\begin{array}{l}
\Delta ABD;AB = 8cm;AD = 6cm\\
\Rightarrow BD = \sqrt {A{B^2} + A{D^2}} = 10cm
\end{array}$
Mặt khác:
$\begin{array}{l}
\Delta ABH \sim \Delta BDC\\
\Rightarrow \dfrac{{AH}}{{BC}} = \dfrac{{AB}}{{BD}} = \dfrac{{BH}}{{DC}}\\
\Rightarrow \left\{ \begin{array}{l}
AH = \dfrac{{BC.AB}}{{BD}} = \dfrac{{6.8}}{{10}} = 4,8cm\\
BH = \dfrac{{DC.AB}}{{BD}} = \dfrac{{{8^2}}}{{10}} = 6,4cm
\end{array} \right.
\end{array}$
Và:
$\begin{array}{l}
\dfrac{{EB}}{{ED}} = \dfrac{{CB}}{{CD}} = \dfrac{6}{8} = \dfrac{3}{4}\\
\Rightarrow EB = \dfrac{3}{7}BD = \dfrac{{30}}{7}cm\\
\Rightarrow EH = BH - EB = 6,4 - \dfrac{{30}}{7} = \dfrac{{74}}{{35}}cm
\end{array}$
Thay $EH = \dfrac{{74}}{{35}}cm;AH = 4,8cm$ vào $(*)$ ta có:
${S_{AECH}} = 4,8.\dfrac{{74}}{{35}} = \dfrac{{1776}}{{175}} \approx 10,15c{m^2}$
Vậy ${S_{AECH}} \approx 10,15c{m^2}$
