Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\Delta ABC;AB = 5cm;AC = 12cm\\
\Rightarrow BC = \sqrt {A{B^2} + A{C^2}} = 13cm
\end{array}$
Vậy $BC=13cm$
b) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BAD} = \widehat {BHD} = {90^0}\\
BDchung\\
\widehat {ABD} = \widehat {HBD}
\end{array} \right.\\
\Rightarrow \Delta ABD = \Delta HBD\left( {ch - gn} \right)
\end{array}$
c) Ta có:
$\begin{array}{l}
\Delta ABD = \Delta HBD\left( {ch - gn} \right)\\
\Rightarrow AD = HD
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {DAK} = \widehat {DHC} = {90^0}\\
AD = HD\\
\widehat {ADK} = \widehat {HDC}\left( {dd} \right)
\end{array} \right.\\
\Rightarrow \Delta ADK = \Delta HDC\left( {g.c.g} \right)\\
\Rightarrow AK = HC\left( 1 \right)
\end{array}$
Lại có
$\begin{array}{l}
\Delta HDC;\widehat H = {90^0}\\
\Rightarrow CD > HC\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow AK < CD$
d) Ta có:
$\begin{array}{l}
\Delta ABD = \Delta HBD\left( {ch - gn} \right)\\
\Rightarrow \left\{ \begin{array}{l}
AD = HD\\
AB = HB
\end{array} \right.\\
\Rightarrow BD \text{là trung trực của AH}
\end{array}$
