Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
DE//AB\\
\Rightarrow \left\{ \begin{array}{l}
\widehat {EDC} = \widehat {ABC}\\
\widehat {ADE} = \widehat {BAD} \Rightarrow \widehat {ADE} = \widehat {EAD} \Rightarrow \Delta AED \text{cân ở E} \Rightarrow EA = ED\left( 1 \right)
\end{array} \right.
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Cchung\\
\widehat {EDC} = \widehat {ABC}
\end{array} \right.\\
\Rightarrow \Delta EDC \sim \Delta ABC\left( {g.g} \right)\\
\Rightarrow \dfrac{{ED}}{{AB}} = \dfrac{{EC}}{{AC}}\left( 2 \right)\\
\text{Từ} \left( 1 \right),\left( 2 \right) \Rightarrow \dfrac{{EA}}{{AB}} = \dfrac{{EC}}{{AC}}\\
\Rightarrow AC.EA = AB.EC
\end{array}$
b) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {API} = \widehat {DPE}\left( {dd} \right)\\
\widehat {IAP} = \widehat {EDP}\left( {AI//DE} \right)
\end{array} \right.\\
\Rightarrow \Delta API \sim \Delta DPE\left( {g.g} \right)\\
\Rightarrow \dfrac{{PI}}{{PE}} = \dfrac{{AI}}{{DE}}\left( 3 \right)
\end{array}$
Chứng minh tương tự có:
$\begin{array}{l}
\Delta BQI \sim \Delta EQD\left( {g.g} \right)\\
\Rightarrow \dfrac{{QI}}{{QD}} = \dfrac{{BI}}{{ED}}\left( 4 \right)\\
\text{Từ} \left( 3 \right),\left( 4 \right) \Rightarrow \dfrac{{PI}}{{PE}} = \dfrac{{QI}}{{QD}}\left( {do:AI = BI} \right)\\
\text{Và} \Rightarrow \dfrac{{PI}}{{IE}} = \dfrac{{QI}}{{ID}}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Ichung\\
\dfrac{{IP}}{{IE}} = \dfrac{{IQ}}{{ID}}
\end{array} \right.\\
\Rightarrow \Delta IPQ \sim \Delta IED\left( {c.g.c} \right)
\end{array}$
