Đáp án:
$\begin{array}{l}
1)AH.BC = AB.AC\\
2)A{H^2} = BH.CH\\
3)A{B^2} = BH.BC\\
4)A{C^2} = CH.BC\\
5)\frac{1}{{A{H^2}}} = \frac{1}{{A{B^2}}} + \frac{1}{{A{C^2}}}\\
+ )\sin B = c{\rm{osC}} = \frac{{AC}}{{BC}};\\
+ )c{\rm{osB = sinC = }}\frac{{AB}}{{BC}}\\
+ )\tan B = \cot C = \frac{{AC}}{{AB}}\\
+ )\cot B = \tan C = \frac{{AB}}{{AC}}
\end{array}$