Giải thích các bước giải:
a.Xét $\Delta AHB, \Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to \Delta ABH\sim\Delta CBA(g.g)$
$\to \dfrac{AB}{BC}=\dfrac{BH}{BA}$
$\to AB^2=BH.BC=36$
$\to AB=6$
$\to AH=\sqrt{AB^2-BH^2}=\dfrac{24}{5}, AC=\sqrt{BC^2-AB^2}=8$
b.Xét $\Delta AHD, \Delta CHE$ có:
$\widehat{AHD}=90^o-\widehat{AHE}=\widehat{EHC}$
$\widehat{DAH}=90^o-\widehat{HAC}=\widehat{HCA}=\widehat{HCE}$
$\to \Delta AHD\sim\Delta CHE(g.g)$
c.Từ câu b
$\to \dfrac{AD}{CE}=\dfrac{AH}{CH}$
Xét $\Delta AHC, \Delta ABC$ có:
Chung $\hat C$
$\widehat{CHA}=\widehat{CAB}(=90^o)$
$\to \Delta CHA\sim\Delta CAB(g.g)$
$\to \dfrac{HA}{AB}=\dfrac{CH}{CA}$
$\to \dfrac{AH}{CH}=\dfrac{AB}{AC}$
$\to \dfrac{AD}{CE}=\dfrac{AB}{AC}$
$\to AD.AC=EC.AB$
d.Gọi $AD\cap HE=F$
Xét $\Delta FDH, \Delta FAE$ có:
Chung $\hat F$
$\widehat{FHD}=\widehat{FAE}(=90^o)$
$\to \Delta FDH\sim\Delta FEA(g.g)$
$\to \dfrac{FD}{FE}=\dfrac{FH}{FA}$
$\to \dfrac{FD}{FH}=\dfrac{FE}{FA}$
Mà $\widehat{AFH}=\widehat{DFE}$
$\to \Delta FAH\sim\Delta FED(c.g.c)$
Gọi $AH\cap DE=G$
$\to \widehat{DAG}=\widehat{GEH}$
Mà $\widehat{AGD}=\widehat{EGH}$
$\to \Delta GAD\sim\Delta GEH(g.g)$
$\to \widehat{GDA}=\widehat{GHE}=90^o-\widehat{GHD}=90^o-\dfrac12\widehat{AHB}=45^o$
$\to \widehat{ADE}=45^o$
Mà $\widehat{DAE}=90^o$
$\to \Delta ADE$ vuông cân tại $A$
$\to AD=AE$
