Giải thích các bước giải:
a.Xét $\Delta AHC,\Delta ABC$ có:
chung $\hat C$
$\widehat{AHC}=\widehat{CAB}(=90^o)$
$\to \Delta ACH\sim\Delta BCA(g.g)$
b.Xét $\Delta AHB,\Delta AHC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to\Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC$
c. Ta có $HD, HE$ là phân giác $\widehat{AHB},\widehat{AHC}$
$\to HD\perp HE$
Gọi $AB\cap HE=F$
Xét $\Delta FDH,\Delta FEA$ 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 $\widehat{AFH}=\widehat{DFE}$
$\to\Delta FDE\sim\Delta FAH(c.g.c)$
$\to \widehat{FDE}=\widehat{FHA}=180^o-\widehat{AHE}=180^o-\dfrac12\widehat{AHC}=180^o-\dfrac12\cdot 90^o=135^o$
$\to \widehat{ADE}=180^o-\widehat{FDE}=45^o$
$\to \Delta ADE$ vuông cân tại $A$
$\to AD=AE$
