$a)$Xét $\Delta BEM$ và $\Delta CFM$
$\widehat{BEM}=\widehat{CFM}=90^o\\ BM=CM\\ \widehat{M_1}=\widehat{M_2}(đđ)\\ \Rightarrow \Delta BEM = \Delta CFM\\ \Rightarrow BE=CF\\ BE \perp AM;CF \perp AM\\ \Rightarrow BE //CF$
$b)$Xét $\Delta BEM$ và $\Delta CFM$
$\widehat{BEM}=\widehat{CFM}=90^o\\ BE=CF\\ BM=CM\\ \Rightarrow \Delta BEM = \Delta CFM\\ \Rightarrow EM=FM\\ AE+AF=AE+AM+MF=AE+AM+EM=2AM\\ c)AM=MC=\dfrac{1}{2}BC$
$\Rightarrow \Delta AMB,AMC$ cân tại $M$
$\Rightarrow \widehat{ACM}=\widehat{MAC};\widehat{ABM}=\widehat{MAB}\\ \Delta ABC:\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^o\\ \Leftrightarrow \widehat{ABC}+\widehat{BAC}+\widehat{ACM}=180^o\\ \Leftrightarrow \widehat{ABC}+\widehat{MAC}+\widehat{MAB}+\widehat{ACM}=180^o\\ \Leftrightarrow 2\widehat{MAC}+2\widehat{MAB}=180^o\\ \Leftrightarrow \widehat{MAC}+\widehat{MAB}=90^o\\ \Leftrightarrow \widehat{BAC}=90^o$
$d)$Xét $\Delta EMC$ và $\Delta FMB$
$\widehat{EMC}=\widehat{FMB}(đđ)\\ MC=MB\\ EM=FM\\ \Rightarrow \Delta EMC=\Delta FMB\\ \Rightarrow \widehat{ECM}=\widehat{FBM}\\ \Rightarrow EC//BF$
Mà $MI \perp EC,MK \perp BK$
$\Rightarrow MI//MK$
$\Rightarrow M,I,K$ thẳng hàng(Do $2$ đường có chung điểm $M)$
