Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Achung\\
\widehat {AHB} = \widehat {ABE} = {90^0}
\end{array} \right.\\
\Rightarrow \Delta HBA \sim \Delta BEA\left( {g.g} \right)\\
\Rightarrow \dfrac{{BA}}{{EA}} = \dfrac{{HA}}{{BA}}\\
\Rightarrow B{A^2} = HA.EA\\
\Rightarrow A{B^2} = AH.AE
\end{array}$
b) Ta có:
$\begin{array}{l}
\Delta HBA \sim \Delta BEA\left( {g.g} \right)\\
\Rightarrow \widehat {HBA} = \widehat {BEA}\\
\Rightarrow \widehat {HBA} = \widehat {HEB}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {HBA} = \widehat {HEB}\\
\widehat {AHB} = \widehat {BHE} = {90^0}
\end{array} \right.\\
\Rightarrow \Delta HAB \sim \Delta HBE\left( {g.g} \right)\\
\Rightarrow \dfrac{{HB}}{{HE}} = \dfrac{{HA}}{{HB}}\\
\Rightarrow H{B^2} = HA.HE
\end{array}$
c) Ta có:
$\begin{array}{l}
\widehat {DFH} = \widehat {BAE} = \widehat {EBH}\left( { + \widehat {AEB} = {{90}^0}} \right)\\
\Rightarrow \widehat {DFH} = \widehat {EBH}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {DFH} = \widehat {EBH}\\
\widehat {DHF} = \widehat {EHB} = {90^0}
\end{array} \right.\\
\Rightarrow \Delta DHF \sim \Delta EHB\left( {g.g} \right)\\
\Rightarrow \dfrac{{HF}}{{HB}} = \dfrac{{HD}}{{HE}}\\
\Rightarrow HE.HF = HB.HD\left( 1 \right)
\end{array}$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHD} = \widehat {BHA} = {90^0}\\
\widehat {ADH} = \widehat {BAH}\left( { + \widehat {HAD} = {{90}^0}} \right)
\end{array} \right.\\
\Rightarrow \Delta AHD \sim \Delta BHA\left( {g.g} \right)\\
\Rightarrow \dfrac{{AH}}{{BH}} = \dfrac{{HD}}{{HA}}\\
\Rightarrow A{H^2} = HB.HD\left( 2 \right)\\
\text{ Từ} \left( 1 \right),\left( 2 \right) \Rightarrow A{H^2} = HE.HF
\end{array}$
