Giải thích các bước giải:
a.Xét $\Delta HAB, \Delta HAC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{HCA}$
$\to \Delta HAB\sim\Delta HCA(g.g)$
b.Xét $\Delta AHB , \Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to \Delta BAH\sim\Delta BCA(g.g)$
$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to BA^2=BH.BC$
Mà $BA=BE(=6)$
$\to BE^2=BH.BC$
c.Ta có $S_{ABC}=\dfrac12AB\cdot AC=24$
Vì $BD$ là phân giác $\hat B$
$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$
$\to \dfrac{DA}{DA+DC}=\dfrac3{3+5}$
$\to \dfrac{DA}{AC}=\dfrac38$
$\to AD=\dfrac38AC$
$\to AD=3$
$\to S_{ABD}=\dfrac12AB\cdot AC=9$
Xét $\Delta ABD, \Delta DBE$ có:
chung $BD$
$\widehat{ABD}=\widehat{EBD}$
$BA=BE$
$\to \Delta ABD=\Delta EBD(c.g.c)$
$\to S_{ABD}=S_{EBD}$
$\to S_{DEC}=S_{ABC}-2S_{ABD}=6$
