Giải thích các bước giải:
Xét tứ giác $AKCH$ có:
$\widehat {KAH} + \widehat {AHC} + \widehat {HCK} + \widehat {CKA} = {360^0}$
$\begin{array}{l}
\Rightarrow \widehat {KAH} + \widehat {HCK} = {180^0}\left( {do:\widehat {AHC} = \widehat {CKA} = {{90}^0}} \right)\\
\Rightarrow \widehat {KAH} + \widehat {DCB} = {180^0}\\
\Rightarrow \widehat {KAH} = {180^0} - \widehat {DCB}\\
\Rightarrow \widehat {KAH} = \widehat {ABC}\left( {do:AB//CD} \right)
\end{array}$
Lại có:
$\begin{array}{l}
{S_{ABCD}} = AH.CD = AK.BC\\
\Rightarrow \dfrac{{AH}}{{BC}} = \dfrac{{AK}}{{CD}}\\
\Rightarrow \dfrac{{AH}}{{BC}} = \dfrac{{AK}}{{BA}}\left( {do:BA = CD} \right)
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {KAH} = \widehat {ABC}\\
\dfrac{{AH}}{{BC}} = \dfrac{{AK}}{{BA}}
\end{array} \right.\\
\Rightarrow \Delta KAH \sim \Delta ABC\left( {c.g.c} \right)
\end{array}$
