Đáp án: $BH = \dfrac{9}{2}cm;CH = 8cm$
Giải thích các bước giải:
$\begin{array}{l}
AB:AC = 3:4\\
\Leftrightarrow \left\{ \begin{array}{l}
AB = 3k\\
AC = 4k
\end{array} \right.\left( {k > 0} \right)
\end{array}$
Theo hệ thức lượng trong tam giác vuông ta có:
$\begin{array}{l}
\dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}} = \dfrac{1}{{A{H^2}}}\\
\Leftrightarrow \dfrac{1}{{{{\left( {3k} \right)}^2}}} + \dfrac{1}{{{{\left( {4k} \right)}^2}}} = \dfrac{1}{{{6^2}}}\\
\Leftrightarrow \dfrac{1}{{{k^2}}}.\left( {\dfrac{1}{9} + \dfrac{1}{{16}}} \right) = \dfrac{1}{{{6^2}}}\\
\Leftrightarrow {k^2} = \dfrac{{25}}{4}\\
\Leftrightarrow k = \dfrac{5}{2}\\
\Leftrightarrow \left\{ \begin{array}{l}
AB = \dfrac{{15}}{2}\left( {cm} \right)\\
AC = 10\left( {cm} \right)
\end{array} \right.\\
Do:A{B^2} = A{H^2} + B{H^2}\\
\Leftrightarrow BH = \sqrt {{{\left( {\dfrac{{15}}{2}} \right)}^2} - {6^2}} = \dfrac{9}{2}\left( {cm} \right)\\
CH = \sqrt {A{C^2} - A{H^2}} = 8\left( {cm} \right)\\
Vậy\,BH = \dfrac{9}{2}cm;CH = 8cm
\end{array}$
