a)

Xét $\Delta ABH$ và $\Delta ACK$, ta có:

+   $\widehat{A}$ là góc chung

+   $\widehat{AHB}=\widehat{AKC}=90{}^\circ $

$\to \Delta ABH\backsim\Delta ACK\left( g.g \right)$

b)

Vì $\Delta ABH\backsim\Delta ACK\left( cmt \right)$

$\to \dfrac{AB}{AC}=\dfrac{AH}{AK}$

$\to \dfrac{AH}{AB}=\dfrac{AK}{AC}$

Xét $\Delta AHK$ và $\Delta ABC$, ta có:

+   $\widehat{A}$ là góc chung

+   $\dfrac{AH}{AB}=\dfrac{AK}{AC}\left( cmt \right)$

$\to \Delta AHK\backsim\Delta ABC\left( c.g.c \right)$

c)

Vì $\Delta AHK\backsim\Delta ABC\left( cmt \right)$

$\to \dfrac{{{S}_{\Delta AHK}}}{{{S}_{\Delta ABC}}}={{\left( \dfrac{HK}{BC} \right)}^{2}}$

$\to \dfrac{1}{9}={{\left( \dfrac{HK}{BC} \right)}^{2}}$

$\to \dfrac{1}{3}=\dfrac{HK}{BC}$

$\to \dfrac{1}{3}=\dfrac{HK}{12}$

$\to HK=\dfrac{12.1}{3}$

$\to HK=4cm$