Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
MA = MK\\
\widehat {AMB} = \widehat {KMC}\left( {dd} \right)\\
MB = MC
\end{array} \right.\\
\Rightarrow \Delta MAB = \Delta MKC\left( {c.g.c} \right)
\end{array}$
b) Ta có;
$\begin{array}{l}
\Delta MAB = \Delta MKC\left( {c.g.c} \right)\\
\Rightarrow \widehat {MBA} = \widehat {MCK}\\
\Rightarrow AB//KC
\end{array}$
Lại có:
$AB \bot AC \Rightarrow KC \bot AC \Rightarrow \widehat {ACK} = {90^0}$
c) Ta có:
$\begin{array}{l}
\Delta MAB = \Delta MKC\left( {c.g.c} \right)\\
\Rightarrow AB = KC
\end{array}$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
IA = IC\\
\widehat {IAB} = \widehat {ICK} = {90^0}\\
AB = CK
\end{array} \right.\\
\Rightarrow \Delta IAB = \Delta ICK\left( {c.g.c} \right)\\
\Rightarrow IB = IK
\end{array}$
d) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
ACchung\\
\widehat {ACK} = \widehat {CAB} = {90^0}\\
CK = AB
\end{array} \right.\\
\Rightarrow \Delta ACK = \Delta CAB\left( {c.g.c} \right)\\
\Rightarrow AK = CB\\
\Rightarrow 2MK = 2MB\\
\Rightarrow MK = MB
\end{array}$
Khi đó:
$\left\{ \begin{array}{l}
MK = MB\\
IK = IB
\end{array} \right.$
$ \Rightarrow IM$ là trung trực của $BK$
$\to IM\bot BK$
$\to IE\bot BK$
