Giải thích các bước giải:

1.Xét $\Delta HBA,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}(=90^o)$

$\to \Delta HBA\sim\Delta ABC(g.g)$

2.Xét $\Delta BIH,\Delta BEC$ có:

Chung $\hat B$

$\hat H=\hat E(=90^o)$

$\to \Delta BHI\sim\Delta BEC(g.g)$

$\to \dfrac{BH}{BE}=\dfrac{BI}{BC}$

$\to BI.BE=BH.BC$

Từ 1 $\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$

$\to AB^2=BH.BC$

$\to AB^2=BH.BC=BI.BE$

3.Xét $\Delta HAB,\Delta HAC$ có:

$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAB}=90^o-\widehat{HAC}=\widehat{HCA}$

$\to \Delta HAB\sim\Delta HCA(g.g)$

$\to \dfrac{HA}{HC}=\dfrac{HB}{HA}$

$\to AH^2=HB.HC$

$\to 2AH.\dfrac12AH=HB.HC$

$\to HD.HI=HB.HC$

$\to \dfrac{HD}{HB}=\dfrac{HC}{HI}$

$\to \Delta HBI\sim\Delta HDC(c.g.c)$

$\to \widehat{DCH}=\widehat{BIH}$

Từ 2 $\to \widehat{BIH}=\widehat{BCE}$

$\to \widehat{HCD}=\widehat{BCE}$

$\to C, E, D$ thẳng hàng