Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {ADB} = \widehat {AEC} = {90^0}\\
\widehat Achung
\end{array} \right.\\
\Rightarrow \Delta ABD \sim \Delta ACE\left( {g.g} \right)\\
\Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{AD}}{{AE}}
\end{array}$
$ \Rightarrow AB.AE = AC.AD$
b) Ta có:
$\begin{array}{l}
\Delta ABD \sim \Delta ACE\left( {g.g} \right)\\
\Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{AD}}{{AE}} \Rightarrow \dfrac{{AB}}{{AD}} = \dfrac{{AC}}{{AE}}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{AB}}{{AD}} = \dfrac{{AC}}{{AE}}\\
\widehat Achung
\end{array} \right.\\
\Rightarrow \Delta ABC \sim \Delta ADE\left( {c.g.c} \right)
\end{array}$
c) Ta có:
$\begin{array}{l}
\Delta ABC \sim \Delta ADE\left( {c.g.c} \right)\\
\Rightarrow \widehat {ACB} = \widehat {AED}\\
\Rightarrow \widehat {ICD} = \widehat {AED}\\
\Rightarrow \widehat {ICD} = \widehat {IEB}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Ichung\\
\widehat {ICD} = \widehat {IEB}
\end{array} \right.\\
\Rightarrow \Delta IDC \sim \Delta IBE\left( {g.g} \right)
\end{array}$
d) Ta có:
$\begin{array}{l}
\Delta IDC \sim \Delta IBE\left( {g.g} \right)\\
\Rightarrow \dfrac{{ID}}{{IB}} = \dfrac{{IC}}{{IE}}\\
\Rightarrow ID.IE = IB.IC\\
\Rightarrow ID.IE = \left( {IO - BO} \right)\left( {IO + OC} \right)\\
\Rightarrow ID.IE = I{O^2} - O{C^2}
\end{array}$
