Giải thích các bước giải:

a.Xét $\Delta ABC,\Delta HBA$ có:

Chung $\hat B$

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

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

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

$\to AB^2=BH.BC$

b.Xét $\Delta ABD,\Delta HBE$ có:

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

$\widehat{ABD}=\widehat{HBE}$

$\to \Delta ABD\sim\Delta HBE(g.g)$

$\to \dfrac{AD}{HE}=\dfrac{BD}{BE}$

$\to HE.BD=BE.AD$

c.Vì $BK\perp MC, CA\perp MB, AC\cap BK=D$

$\to D$ là trực tâm $\Delta MBC$

$\to MD\perp BC$

Xét $\Delta MKB,\Delta MAC$ có:

Chung $\hat M$

$\widehat{MKB}=\widehat{MAC}(=90^o)$

$\to \Delta MKB\sim\Delta MAC(g.g)$

$\to \dfrac{MK}{MA}=\dfrac{MB}{MC}$

$\to \Delta MKA\sim\Delta MBC(c.g.c)$

$\to \widehat{MAK}=\widehat{MBC}$

Tương tự chứng minh được: $\widehat{CKN}=\widehat{MBC}$

$\to \widehat{MAK}=\widehat{NKC}$

$\to 90^o-\widehat{MKA}=90^o-\widehat{NKC}$

$\to \widehat{BKA}=\widehat{BKN}$

$\to KB$ là phân giác $\widehat{AKN}$