Giải thích các bước giải:
a) Ta có;
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};\widehat C = {35^0}\\
\Rightarrow \widehat B = {90^0} - \widehat C = {55^0}\\
\Rightarrow \widehat B > \widehat C\\
\Rightarrow AC > AB
\end{array}$
b) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
BHchung\\
\widehat {AHB} = \widehat {DHB} = {90^0}\\
AH = DH
\end{array} \right.\\
\Rightarrow \Delta AHB = \Delta DHB\left( {c.g.c} \right)\\
\Rightarrow AB = DB
\end{array}$
$ \Rightarrow \Delta ABD$ cân ở $B$
c) Ta có;
$\begin{array}{l}
\Delta AHB = \Delta DHB\left( {c.g.c} \right)\\
\Rightarrow \widehat {ABH} = \widehat {DBH}\\
\Rightarrow \widehat {ABC} = \widehat {DBC}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
AB = DB\\
\widehat {ABC} = \widehat {DBC}\\
BCchung
\end{array} \right.\\
\Rightarrow \Delta ABC = \Delta DBC\left( {c.g.c} \right)\\
\Rightarrow \widehat {BAC} = \widehat {BDC}\\
\Rightarrow \widehat {BDC} = {90^0}
\end{array}$
d) Ta có:
$\begin{array}{l}
\Delta AHB;\widehat {AHB} = {90^0};AH = 5cm;AB = 13cm\\
\Rightarrow HB = \sqrt {A{B^2} - A{H^2}} = 12cm
\end{array}$
Vậy $HB=12cm$
e) Ta có:
$\begin{array}{l}
\Delta AHB = \Delta DHB\\
\Rightarrow \widehat {HAB} = \widehat {HDB}\left( 1 \right)
\end{array}$
Mà
$\begin{array}{l}
HK//AB\left( { \bot AC} \right)\\
\Rightarrow \widehat {HAB} = \widehat {KHA}\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow \widehat {KHA} = \widehat {HDB} \Rightarrow \widehat {KHA} = \widehat {BDA}$
