1)

$BD$ là phân giác $\widehat{ABC}$

$\to \dfrac{AB}{AD}=\dfrac{BC}{CD}=\dfrac{AB+BC}{AD+CD}=\dfrac{6+4}{6}=\dfrac{5}{3}$

$\dfrac{AB}{AD}=\dfrac{5}{3}\to AD=\dfrac{3}{5}AB=\dfrac{3}{5}.6=3,6\,\,\left( cm \right)$

$BD$ là phân giác $\widehat{ABC}$

$\to \dfrac{AB}{BC}=\dfrac{AD}{CD}$

$CE$ là phân giác $\widehat{ACB}$

$\to \dfrac{AC}{BC}=\dfrac{AE}{BE}$

Mà: $AB=AC$

Nên: $\dfrac{AD}{CD}=\dfrac{AE}{BE}$

$\to ED\,//\,BC$

$\to \dfrac{AD}{AC}=\dfrac{ED}{BC}\,\,\to \,\,ED=\dfrac{AD.BC}{AC}=\dfrac{3,6.4}{6}=2,4\,\,\left( cm \right)$

2)

Xét $\Delta ADB$ và $\Delta AEC$, ta có:

$\widehat{BAC}$ là góc chung

$\widehat{ABD}=\widehat{ACE}\,\,\,\left( =\dfrac{1}{2}\widehat{ABC}=\dfrac{1}{2}\widehat{ACB} \right)$

$\to \Delta ADB\backsim\Delta AEC\,\,\,\left( g.g \right)$

3)

Xét $\Delta IEB$ và $\Delta IDC$, ta có:

$\widehat{EIB}=\widehat{DIC}$ ( hai góc đối đỉnh )

$\widehat{IBE}=\widehat{ICD}\,\,\,\left( =\dfrac{1}{2}\widehat{ABC}=\dfrac{1}{2}\widehat{ACB} \right)$

$\to \Delta IEB\backsim\Delta IDC\,\,\,\left( g.g \right)$

$\to \dfrac{IE}{ID}=\dfrac{BE}{CD}\,\,\to \,\,IE.CD=ID.BE$

4)

Vì $ED\,//\,BC$

$\to \Delta AED\backsim\Delta ABC$

$\to \dfrac{{{S}_{\Delta AED}}}{{{S}_{\Delta ABC}}}={{\left( \dfrac{AD}{AC} \right)}^{2}}={{\left( \dfrac{3,6}{6} \right)}^{2}}=\dfrac{9}{25}$

$\to {{S}_{\Delta AED}}=\dfrac{9}{25}{{S}_{\Delta ABC}}=\dfrac{9}{25}.60=21,6\,\,\left( c{{m}^{2}} \right)$