Đáp án:
b) Xét ΔAMH và ΔAHB có:
+ góc AMH = góc AHB = 90 độ
+ góc MAH chung
=> ΔAMH ~ ΔAHB (g-g)
$\begin{array}{l}
\Rightarrow \dfrac{{AM}}{{AH}} = \dfrac{{AH}}{{AB}}\\
\Rightarrow AM.AB = A{H^2}
\end{array}$
Chứng minh tương tự ta có:
ΔANH ~ ΔAHC (g-g)
$\begin{array}{l}
\Rightarrow \dfrac{{AN}}{{AH}} = \dfrac{{AH}}{{AC}}\\
\Rightarrow AN.AC = A{H^2}
\end{array}$
$\begin{array}{l}
\Rightarrow AM.AB = AN.AC\\
\Rightarrow \dfrac{{AM}}{{AN}} = \dfrac{{AC}}{{AB}}\\
Xét:\Delta AMN;\Delta ACB:\\
+ \dfrac{{AM}}{{AN}} = \dfrac{{AC}}{{AB}}\\
+ \widehat {MAN}\,chung\\
\Rightarrow \Delta AMN \sim \Delta ACB\left( {c - g - c} \right)
\end{array}$
