Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow B{C^2} = {8^2} + {15^2} = 225 + 64 = 289\\
\Rightarrow BC = 17\\
+ AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{8.15}}{{17}} = \dfrac{{120}}{{17}}\\
+ B{H^2} = A{B^2} - A{H^2} = {8^2} - {\left( {\dfrac{{120}}{{17}}} \right)^2}\\
\Rightarrow BH = \dfrac{{64}}{{17}}
\end{array}$
b)
$\begin{array}{l}
\text{Xét}:\Delta AMH;\Delta AHB:\\
+ )\widehat {AMH} = \widehat {AHB} = {90^0}\\
+ \widehat {MAH\,}\,chung\\
\Rightarrow \Delta AMH \sim \Delta AHB\left( {g - g} \right)\\
\Rightarrow \dfrac{{AM}}{{AH}} = \dfrac{{AH}}{{AB}}\\
\Rightarrow AM.AB = A{H^2}\\
\text{Tương}\,\text{tự}:AN.AC = A{H^2}\\
\Rightarrow AM.AB = AN.AC\\
\Rightarrow \Delta AMN \sim \Delta ACB\left( {c - g - c} \right)\\
\Rightarrow \dfrac{{{S_{AMN}}}}{{{S_{ACB}}}} = {\left( {\dfrac{{MN}}{{BC}}} \right)^2}\\
= {\left( {\dfrac{{AH}}{{BC}}} \right)^2} = \dfrac{{{{120}^2}}}{{{{289}^2}}}\\
\Rightarrow {S_{AMN}} = \dfrac{{{{120}^2}}}{{{{289}^2}}}.\dfrac{1}{2}.AB.AC = 10,344
\end{array}$
