Giải thích các bước giải:
a.Xét $\Delta ABC,\Delta HBA$ có:
Chung $\hat B$
$\widehat{BAC}=\widehat{AHB}(=90^o)$
$\to\Delta ABC\sim\Delta HBA(g.g)$
b.Từ câu a
$\to \dfrac{AB}{HB}=\dfrac{BC}{BA}$
$\to AB^2=BH.BC$
$\to BH=\dfrac{AB^2}{BC}=\dfrac95$
c. Ta có $BE$ là phân giác $\hat B$
$\to \dfrac{EH}{EA}=\dfrac{BH}{BA}=\dfrac35$
$\to \dfrac{EH}{EH+EA}=\dfrac3{3+5}$
$\to \dfrac{EH}{AH}=\dfrac38$
$\to EH=\dfrac38AH$
Mà $AH=\sqrt{AB^2-BH^2}=\dfrac{12}5$
$\to EH=\dfrac9{10}$
d.Ta có $\Delta ABC$ vuông tại $A\to AC=\sqrt{BC^2-AB^2}=4$
$\to S_{ABC}=\dfrac12AB\cdot AC=6$
Mà $BD$ là phân giác $\hat B$
$\to \dfrac{DC}{DA}=\dfrac{BC}{BA}=\dfrac53$
$\to \dfrac{DC}{DC+DA}=\dfrac5{5+3}$
$\to \dfrac{DC}{AC}=\dfrac58$
$\to \dfrac{S_{BCD}}{S_{BAC}}=\dfrac58$
$\to S_{BCD}=\dfrac58S_{ABC}=\dfrac{15}4$
$\to S_{HEDC}=S_{BCD}-S_{BHE}$
$\to S_{HEDC}=\dfrac{15}4-\dfrac12BH\cdot HE$
$\to S_{HEDC}=\dfrac{15}4-\dfrac{81}{100}$
$\to S_{HEDC}=\dfrac{147}{50}$
