
a) \(\text{Xét }\Delta ABK\text{ và }\Delta BCI\text{ có:}\)
\(AB=BC\left(\text{Hình vuông ABCD}\right)\)
\(AK=BI\left(=\frac{AB}{2}=\frac{AD}{2}\right)\)
\(\Rightarrow\Delta ABK=\Delta BCI\left(\text{Hai cạnh góc vuông}\right)\)
\(\Rightarrow\widehat{AKB}=\widehat{BIC}\left(\text{Hai góc tương ứng}\right)\)
\(\text{Mà }\widehat{AIK}+\widehat{ABK}=90^0\left(\text{phụ nhau}\right)\)
\(\Rightarrow\widehat{IEB}=90^0\)
\(\Rightarrow BK\text{⊥}CI\) (đpcm)