a,

$\Delta AHB$ và $\Delta CAB$ có:

$\widehat{AHB}=\widehat{CAB}=90^o$

$\widehat{B}$ chung 

$\to \Delta AHB\backsim\Delta CAB$ (g.g)

b,

$\Delta AHB\backsim\Delta CAB$ nên:

$\dfrac{AB}{BH}=\dfrac{BC}{AB}$

$\to BH=\dfrac{AB^2}{BC}=9(cm)$

$\to \dfrac{BH}{AB}=\dfrac{3}{4}$

$\Delta HEB$ và $\Delta ADB$ có:

$\widehat{HBE}=\widehat{ABD}$

$\widehat{BHE}=\widehat{BAD}=90^o$

$\to\Delta HEB\backsim\Delta ABD$ (g.g)

Vậy $\dfrac{S_{HEB}}{S_{ABD}}=\left( \dfrac{3}{4}\right)^2=\dfrac{9}{16}$

c,

Theo tính chất phân giác:

$\dfrac{EH}{EA}=\dfrac{BH}{BA}$

$\dfrac{DA}{DC}=\dfrac{AB}{BC}$

Mà $\dfrac{BH}{BA}=\dfrac{BA}{BC}$ nên $\dfrac{EH}{EA}=\dfrac{DA}{DC}$

Vậy $EA.DA=EH.CD$

d,

Kẻ tia $Cx//AB, AF\cap Cx=G$

$\Delta CAG$ và $\Delta ABM$ có:

$\widehat{GCA}=\widehat{MAB}=90^o$

$\widehat{GAC}=\widehat{MBA}=90^o-\widehat{AMB}$

$\to \Delta CAG\backsim\Delta ABM$ (g.g)

$\to \dfrac{GC}{AM}=\dfrac{AC}{AB}=1$

$\to CG=AM=\dfrac{1}{2}AC=\dfrac{1}{2}AB$

$\to \dfrac{GC}{AB}=\dfrac{1}{2}$

Theo Talet: $\dfrac{GC}{AB}=\dfrac{FC}{FB}=\dfrac{1}{2}$

Vậy $BF=2CF$