a, Vì \(\left\{{}\begin{matrix}AD=AB\\AI=DK\left(\dfrac{1}{2}AD=\dfrac{1}{2}DC\right)\\\widehat{BAD}=\widehat{ADK}=90^0\end{matrix}\right.\) nên \(\Delta AIB=\Delta DKA\left(c.g.c\right)\)
\(\Rightarrow\widehat{ABI}=\widehat{DAI}\\ \Rightarrow\widehat{DAI}+\widehat{AIB}=\widehat{ABI}+\widehat{AIB}=90^0\\ \Rightarrow BI\perp AK\)