Giải thích các bước giải:

a.Xét $\Delta ABC,\Delta HAB$ có:

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

$\widehat{ABC}=\widehat{HAB}$ vì $AH//BC$ 

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

b.Xét $\Delta AHB,\Delta KAC$ có:

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

$\widehat{HAB}=180^o-\widehat{BAC}-\widehat{CAK}=90^o-\widehat{CAK}=\widehat{ACK}$

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

$\to \dfrac{HA}{KC}=\dfrac{HB}{KA}$

$\to AH\cdot AK=BH\cdot CK$

c.Từ câu a

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

$\to BH=\dfrac{AB\cdot AC}{BC}=\dfrac{12}5$

$\to AH=\sqrt{AB^2-BH^2}=\dfrac{9}5$

$\to \dfrac{MA}{MB}=\dfrac{AH}{BC}=\dfrac9{25}$

$\to \dfrac{MA+MB}{MB}=\dfrac{9+25}{25}$

$\to \dfrac{AB}{MB}=\dfrac{34}{25}$

$\to \dfrac{S_{ABC}}{S_{MBC}}=\dfrac{34}{25}$

$\to S_{MBC}=\dfrac{25}{34}S_{ABC}=\dfrac{25}{34}\cdot \dfrac12AB\cdot AC=\dfrac{75}{17}$