Đáp án:
a)
$\begin{array}{l}
Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow B{C^2} = {6^2} + {8^2} = 100\\
\Rightarrow BC = 10\left( {cm} \right)\\
b)Xét:\Delta ABC;\Delta HBA:\\
+ \widehat {ABC}\,chung\\
+ \widehat {BAC} = \widehat {BHA} = {90^0}\\
\Rightarrow \Delta ABC \sim \Delta HBA\left( {g - g} \right)\\
c)\Delta ABC \sim \Delta HBA\\
\Rightarrow \frac{{AB}}{{BH}} = \frac{{BC}}{{AB}}\\
\Rightarrow A{B^2} = BH.BC\\
d)Xét:\Delta ABC;\Delta HAC:\\
+ \widehat {ACB}\,chung\\
+ \widehat {BAC} = \widehat {CHC} = {90^0}\\
\Rightarrow \Delta ABC \sim \Delta HAC\left( {g - g} \right)\\
\Rightarrow \frac{{AC}}{{CH}} = \frac{{BC}}{{AC}}\\
\Rightarrow A{C^2} = CH.BC\\
e)Xét:\Delta ABH;\Delta CAH:\\
+ \widehat {ABH} = \widehat {CAH}\\
+ \widehat {AHB} = \widehat {CHA} = {90^0}\\
\Rightarrow \Delta ABH \sim \Delta CAH\left( {g - g} \right)\\
\Rightarrow \frac{{AH}}{{CH}} = \frac{{BH}}{{AH}}\\
\Rightarrow A{H^2} = BH.CH
\end{array}$
