Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHN} = \widehat {CKN} = {90^0}\\
AN = CN\\
\widehat {ANH} = \widehat {CNK}\left( {dd} \right)
\end{array} \right.\\
 \Rightarrow \Delta AHN = \Delta CKN\left( {ch - gn} \right)\\
 \Rightarrow AH = CK
\end{array}$

b) Kẻ $BD\bot MN=D$

Ta có:

$\begin{array}{l}
\Delta AHN = \Delta CKN\left( {ch - gn} \right)\\
 \Rightarrow HN = KN\left( 1 \right)
\end{array}$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHM} = \widehat {BDM} = {90^0}\\
AM = BM\\
\widehat {AMH} = \widehat {BMD}\left( {dd} \right)
\end{array} \right.\\
 \Rightarrow \Delta AHM = \Delta BDM\left( {ch - gn} \right)\\
 \Rightarrow MH = MD\left( 2 \right);AH = BD\\
\text{Từ} \left( 1 \right),\left( 2 \right) \Rightarrow MH + NH = MD + NK\\
 \Rightarrow MN = MD + NK\\
 \Rightarrow 2MN = MN + MD + NK\\
 \Rightarrow 2MN = DK\left( * \right)
\end{array}$

Mặt khác:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {DKC} = \widehat {CBD} = {90^0}\\
CDchung\\
BD = CK\left( { = AH} \right)
\end{array} \right.\\
 \Rightarrow \Delta DKC = \Delta CBD\left( {ch - cgv} \right)\\
 \Rightarrow DK = CB\left( {**} \right)
\end{array}$

Từ $\left( * \right),\left( {**} \right) \Rightarrow BC = 2MN$

Xin hay nhất ạ ^_^