Giải thích các bước giải:
a) Sửa đề: $\Delta HEB \sim \Delta HDC$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BHE} = \widehat {CHD}\left( {dd} \right)\\
\widehat {BEH} = \widehat {CDH} = {90^0}
\end{array} \right.\\
\Rightarrow \Delta HEB \sim \Delta HDC\left( {g.g} \right)\\
\Rightarrow \dfrac{{HE}}{{HD}} = \dfrac{{HB}}{{HC}}\\
\Rightarrow \dfrac{{HD}}{{HC}} = \dfrac{{HE}}{{HB}}
\end{array}$
b) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {DHE} = \widehat {CHB} = {90^0}\\
\dfrac{{HD}}{{HC}} = \dfrac{{HE}}{{HB}}
\end{array} \right.\\
\Rightarrow \Delta HDE \sim \Delta HCB\left( {g.g} \right)
\end{array}$
c) Ta có:
$\begin{array}{l}
\Delta BDC;\widehat {BDC} = {90^0};BD = CD\\
\Rightarrow \Delta BDC\text{vuông cân ở D}\\
\Rightarrow \widehat {DBC} = \widehat {DCB} = {45^0}
\end{array}$
Lại có:
$\begin{array}{l}
\Delta HDE \sim \Delta HCB\left( {g.g} \right)\\
\Rightarrow \widehat {HED} = \widehat {HBC}\\
\Rightarrow \widehat {HED} = {45^0}
\end{array}$
Mặt khác:
$\begin{array}{l}
\widehat {BEH} + \widehat {BMH} = {90^0} + {90^0} = {180^0}\\
\Rightarrow BEHM \text{là tứ giác nội tiếp}\\
\Rightarrow \widehat {HEM} = \widehat {HBM}\\
\Rightarrow \widehat {HEM} = {45^0}
\end{array}$
Như vậy:
$\begin{array}{l}
\widehat {DEM} = \widehat {HEM} + \widehat {HED} = {45^0} + {45^0} = {90^0}\\
\Rightarrow DE \bot EM
\end{array}$
d) Ta có:
$BEHM \text{là tứ giác nội tiếp}$
$ \Rightarrow \widehat {CHM} = \widehat {CBE}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {CHM} = \widehat {CBE}\\
\widehat {CMH} = \widehat {CEB} = {90^0}
\end{array} \right.\\
\Rightarrow \Delta CMH \sim \Delta CEB\left( {g.g} \right)\\
\Rightarrow \dfrac{{CM}}{{CE}} = \dfrac{{CH}}{{CB}}\\
\Rightarrow CH.CE = CE.CB
\end{array}$
Ta cũng có:
$\widehat {HDC} + \widehat {HMC} = {90^0} + {90^0} = {180^0}$
$ \Rightarrow HDCM$ là tứ giác nội tiếp
$ \Rightarrow \widehat {BHM} = \widehat {BCD}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BHM} = \widehat {BCD}\\
\widehat {BMH} = \widehat {BDC} = {90^0}
\end{array} \right.\\
\Rightarrow \Delta BMH \sim \Delta BDC\left( {g.g} \right)\\
\Rightarrow \dfrac{{BH}}{{BC}} = \dfrac{{BM}}{{BD}}\\
\Rightarrow BH.BD = BM.BC
\end{array}$
Như vậy: $BH.BD + CH.CE = BM.BC + CM.CB = B{C^2}$
