Giải thích các bước giải:
b.Ta có : $BC^2=AB^2+AC^2=25\to BC=5$
Từ câu a
$\to \dfrac{HB}{AB}=\dfrac{BA}{BC}\to AB^2=BH.BC$
$\to BH=\dfrac{AB^2}{BC}=\dfrac95$
$\to CH=BC-BH=\dfrac{16}5$
c.Ta có: $AH\perp BC, AB\perp AC\to AH.BC=AB.AC(=2S_{ABC})$
$\to AH=\dfrac{AB.AC}{BC}=\dfrac{12}5=2AK\to \dfrac{AM}{AB}=\dfrac{AK}{AH}=\dfrac12$
Mà $MN//BC\to \widehat{AMN}=\widehat{ABC}$
$\to \Delta AMN\sim\Delta ABC(g.g)$
$\to \dfrac{S_{AMN}}{S_{ABC}}=(\dfrac{AM}{AB})^2=\dfrac14$
$\to \dfrac{S_{ABC}-S_{AMN}}{S_{ABC}}=\dfrac{4-1}{4}$
$\to \dfrac{S_{CBMN}}{S_{ABC}}=\dfrac3{4}$
$\to S_{BMNC}=\dfrac34S_{ABC}=\dfrac34\cdot\dfrac12AB\cdot AC=\dfrac92$
