giúp e giải bài này vs ạ

Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta ABC$ có:

Chung $\hat B$

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

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

$\to \dfrac{AB}{BC}=\dfrac{HB}{AB}$

$\to AB^2=BH.BC$

Xét $\Delta AHB,\Delta AHC$ có:

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

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

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

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

$\to AH^2=HB.HC$

b.Ta có: $HM//AC$

$\to \dfrac{MN}{AD}=\dfrac{BM}{BA}=\dfrac{HM}{AC}$

$\to \dfrac{MH}{MH}=\dfrac{AD}{AC}$

c.Xét $\Delta BEA,\Delta BAD$ có:

Chung $\hat B$

$\widehat{BEA}=\widehat{BAD}(=90^o)$

$\to \Delta BEA\sim\Delta BAD(g.g)$

$\to \dfrac{BE}{BA}=\dfrac{BA}{BD}$

$\to AB^2=BE.BD$

$\to BE.BD=BH.BC$

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

$\to \Delta BEH\sim\Delta BCD(c.g.c)$

$\to \widehat{BEH}=\widehat{BCD}=\widehat{BCA}=90^o-\widehat{ABC}=90^o-\widehat{ABH}=\widehat{BAH}$