Giải thích các bước giải:
a.Ta có :
$BC^2=AB^2+AC^2=12^2+16^2=400\to BC=20$
Do AD là phân giác góc A
$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$
$\to \dfrac{DA}{DA+DC}=\dfrac3{3+5}$
$\to \dfrac{DA}{AC}=\dfrac38$
$\to AD=\dfrac38AC=6$
$\to CD=AC-AD=10$
b.Ta có : $AB\perp AC, AH\perp BC$
$\to AB.AC=AH.BC(=2S_{ABC})$
$\to AH=\dfrac{AB.AC}{BC}=\dfrac{12\cdot 16}{20}=9.6$
$\to \dfrac{AK}{AH}=\dfrac{3.6}{9.6}=\dfrac38$
Ta có : $MN//BC\to \widehat{AMN}=\widehat{ABC}$
$\to \dfrac{S_{AMN}}{S_{ABC}}=(\dfrac{AM}{AB})^2=(\dfrac{AK}{AH})^2=\dfrac9{64}$
$\to \dfrac{S_{ABC}-S_{AMN}}{S_{ABC}}=\dfrac{64-9}{64}$
$\to \dfrac{S_{BMNC}}{S_{ABC}}=\dfrac{55}{64}$
$\to S_{BMNC}=\dfrac{55}{64}S_{ABC}$
$\to S_{BMNC}=\dfrac{55}{64}\cdot \dfrac12\cdot AB\cdot AC$
$\to S_{BMNC}=\dfrac{55}{64}\cdot \dfrac12\cdot 12\cdot 16$
$\to S_{BMNC}=\dfrac{165}{2}$
