Đáp án:
$\begin{array}{l}
1)\\
+ \Delta ABC:MN//BC\\
\Rightarrow \frac{{AM}}{{AC}} = \frac{{MN}}{{BC}}\\
\Rightarrow \frac{5}{{4 + 3}} = \frac{y}{7}\\
\Rightarrow y = 5\\
+ \Delta ABC:DE//BC\\
\frac{{AE}}{{AC}} = \frac{{AD}}{{AB}}\\
\Rightarrow \frac{4}{7} = \frac{x}{{x + z}}\\
\frac{{AE}}{{EC}} = \frac{{AD}}{{DB}}\\
\Rightarrow \frac{x}{z} = \frac{4}{3}\\
\Rightarrow z = \frac{3}{4}x\\
\Rightarrow \frac{4}{7} = \frac{x}{{x + \frac{3}{4}x}} \Rightarrow x = 1\\
\Rightarrow z = \frac{3}{4}\\
Vậy\,x = 1;y = 5;z = \frac{3}{4}\\
2)\\
+ \Delta ABC \sim \Delta HBA\left( {Do:\left\{ \begin{array}{l}
\widehat {BAC} = \widehat {BHA} = {90^o}\\
\widehat B\,chung
\end{array} \right.} \right)\\
+ \Delta ABC \sim \Delta HAC\left( {g - g} \right)\\
+ \Delta HBA \sim \Delta HAC\left( {g - g} \right)
\end{array}$
