Ta có:
\(AB=AC\)
\(\Rightarrow\Delta ABC\)là tam giác cân
\(\Rightarrow\widehat{ABC}=\widehat{ACB}\)
Do \(\widehat{ACB}\)và \(\widehat{KCE}\)là 2 góc đối đỉnh
\(\Rightarrow\widehat{ACB}=\widehat{KCE}\)
Xét \(\Delta BDH\)(vuông) và \(\Delta CEK\)(vuông) có:
\(BD=CE\)
\(\widehat{DBH}=\widehat{ECK}\left(=\widehat{ACB}\right)\)
\(\Rightarrow\Delta BDH=\Delta CEK\left(ch.gn\right)\)
\(\Rightarrow HD=EK\)
Ta có:
\(\widehat{DIH}=\widehat{KIE}\)(đối đỉnh)
\(\widehat{DHI}=\widehat{EKI}\)(=90O)
\(\Rightarrow\widehat{HDI}=\widehat{KEI}\)
Xét \(\Delta DHI\)và \(\Delta EKI\)có:
\(\widehat{DHI}=\widehat{EKI}\)
\(HD=EK\)
\(\widehat{HDI}=\widehat{KEI}\)
\(\Rightarrow\Delta DHI=\Delta EKI\left(g.c.g\right)\)
\(\Rightarrow DI=IE\)
Do \(\hept{\begin{cases}DI< DE\\DI=IE\end{cases}}\)
Vậy I là trung điểm DE