a, Vì tam giác ABC cân nên : AB = AC
\(\Rightarrow\dfrac{1}{2}AB=\dfrac{1}{2}AC\\ \Leftrightarrow AN=NB=AM=MC\)
\(Xét\Delta BMCvà\Delta CNBcó:\\ \left\{{}\begin{matrix}MC=BN\left(cmt\right)\\\widehat{MCB}=\widehat{NBC}\left(gt\right)\\BClàcạnhchung\end{matrix}\right.\\ \Rightarrow\Delta BMC=\Delta CNB\left(c-g-c\right)\\ \Rightarrow\widehat{BMC}=\widehat{CNB}\left(haigóctươngứng\right)\)
\(Xét\Delta BNIvà\Delta CMKcó:\\ \left\{{}\begin{matrix}\widehat{CKM}=\widehat{BIN}\left(=90^0\right)\\MC=BN\left(cmt\right)\\\widehat{BNI}=\widehat{CMK}\left(cmt\right)\end{matrix}\right.\\ \Rightarrow\Delta BNI=\Delta CMK\left(ch-gn\right)\\ \Rightarrow CK=BI\left(haicạnhtươngứng\right)\)
b, Theo câu a, ta có :
\(\widehat{KCM}=\widehat{IBN}\left(haigóctươngứng\right)\)
\(Xét\Delta AIBvà\Delta AKCcó:\\ \left\{{}\begin{matrix}AB=AC\left(gt\right)\\\widehat{ACK}=\widehat{AIB}\left(cmt\right)\\BI=CK\left(cmt\right)\end{matrix}\right.\\ \Rightarrow\Delta AIB=\Delta AKC\left(c-g-c\right)\\ \Rightarrow AI=AK\left(haicạnhtươngứng\right)\\ \Rightarrow\Delta AIKcântạiA\)
