Hình bạn tự vẽ nhé <3
a/ Ta có : \(\left\{{}\begin{matrix}AB+AE=AE\\BD+DC=BC\end{matrix}\right.\)
Mà \(\left\{{}\begin{matrix}AB=BD\\AE=DC\end{matrix}\right.\)
\(\Leftrightarrow BE=BC\)
Xét \(\Delta EBD\) và \(\Delta CBA\) có :
\(\left\{{}\begin{matrix}\widehat{EBC}chung\\BD=AB\\BE=BC\end{matrix}\right.\)
\(\Leftrightarrow\Delta EBD=\Delta CBA\left(c-g-c\right)\)
\(\Leftrightarrow\widehat{BED}=\widehat{ÂCB}\left(đpcm\right)\)
b/ \(\Delta EBD=\Delta CBA\left(cmt\right)\)
\(\Leftrightarrow\) \(\widehat{BAC}=\widehat{BDE}\)
Lại có : \(\left\{{}\begin{matrix}\widehat{BAC}+\widehat{EAI}=180^0\\\widehat{BDE}+\widehat{IDC}=180^0\end{matrix}\right.\)
\(\Leftrightarrow\widehat{EAI}=\widehat{IDC}\)
Xét \(\Delta AIE\) và \(\Delta DIC\) có :
\(\left\{{}\begin{matrix}\widehat{AEI}=\widehat{ICD}\\AE=DC\\\widehat{IAE}=\widehat{IDC}\end{matrix}\right.\)
\(\Leftrightarrow\Delta AIE=\Delta DIC\left(g-c-g\right)\)
\(\Leftrightarrow IE=IC\)
c/ Xét \(\Delta BEI\) và \(\Delta BCI\) có :
\(\left\{{}\begin{matrix}BE=BC\\IE=IC\\\widehat{BED}=\widehat{ACB}\end{matrix}\right.\)
\(\Leftrightarrow\Delta BEI=\Delta BCI\left(c-g-c\right)\)
\(\Leftrightarrow\widehat{EBI}=\widehat{IBC}\)
\(\Leftrightarrow BI\) là tia phân giác của \(\widehat{EBC}\left(đpcm\right)\)
