Giải thích các bước giải:
a.Xét $\Delta HAB, \Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to\Delta HBA\sim\Delta ABC(g.g)$
b.Từ câu a
$\to \dfrac{BA}{BC}=\dfrac{HB}{AB}$
$\to BA^2=BH.BC$
Mà $BA=BE\to BE^2=BH.BC$
c.Ta có $\Delta ABC$ vuông tại $A\to BC^2=AB^2+AC^2=25\to BC=5$
Mà $AH\perp BC$
$\to S_{ABC}=\dfrac12AB.AC=\dfrac12AH.BC$
$\to AH=\dfrac{AB.AC}{BC}=\dfrac{12}{5}$
d.Vì $BD$ là phân giác $\hat B$
$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$
$\to \dfrac{DA+DC}{DC}=\dfrac{3+5}{5}$
$\to \dfrac{AC}{CD}=\dfrac85$
$\to CD=\dfrac58AC=\dfrac52$
Xét $\Delta ABD, \Delta EBD$ có:
$BA=BE$
$\widehat{ABD}=\widehat{DBE}$
Chung $BD$
$\to\Delta BAD=\Delta BED(c.g.c)$
$\to \widehat{DEB}=\widehat{DAB}=90^o\to DE\perp BC\to \widehat{CED}=90^o$
Xét $\Delta CDE,\Delta CAB$ có:
Chung $\hat C$
$\widehat{CED}=\widehat{CAB}(=90^o)$
$\to\Delta CDE\sim\Delta CBA(g.g)$
$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CD}{CB})^2=\dfrac14$
