a)
Xét $\Delta ABC$ và $\Delta HBA$, ta có:
$\widehat{BAC}=\widehat{BHA}=90{}^\circ $
$\widehat{ABC}$ là góc chung
$\to \Delta ABC\backsim\Delta HBA\,\,\,\left( g.g \right)$
$\to \dfrac{AB}{BH}=\dfrac{BC}{AB}$
$\to A{{B}^{2}}=BH.BC$
b)
$\Delta ABH$ có $BI$ là phân giác $\widehat{ABH}$
$\to \dfrac{IA}{IH}=\dfrac{AB}{BH}$
$\to IA.BH=IH.AB$
c)
Xét $\Delta AHB$ và $\Delta CHA$, ta có:
$\widehat{AHB}=\widehat{CHA}=90{}^\circ $
$\widehat{HAB}=\widehat{HCA}$ ( cùng phụ $\widehat{ABC}$ )
$\to \Delta AHB\backsim\Delta CHA\,\,\,\left( g.g \right)$
$\to \dfrac{{{S}_{\Delta AHB}}}{{{S}_{\Delta CHA}}}={{\left( \dfrac{AB}{AC} \right)}^{2}}={{\left( \dfrac{6}{8} \right)}^{2}}=\dfrac{9}{16}$