Giải thích các bước giải:

a.Xét $\Delta ABH,\Delta ABC$ có:

Chung $\hat B$

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

$\to \Delta ABH\sim\Delta CBA(g.g)$

$\to \dfrac{AB}{BC}=\dfrac{BH}{BA}$

$\to AB^2=BH\cdot BC$

Xét $\Delta ACH, \Delta ABC$ có:

Chung $\hat A$

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

$\to \Delta CAH\sim\Delta CBA(g.g)$

$\to \dfrac{CA}{CB}=\dfrac{CH}{CA}$

$\to AC^2=HC\cdot BC$

b.Từ a $\to \Delta CAH\sim\Delta ABH(\sim\Delta CBA)$

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

$\to HA^2=HB\cdot HC$

c.Ta có: $ \Delta CAH\sim\Delta ABH, M, N$ là trung điểm $HA, HB$

$\to \Delta CMA\sim\Delta ANB$