Đáp án: $\begin{array}{l}
AB = 10cm;AC = 24cm;AH = \dfrac{{120}}{{13}}cm;\\
BH = \dfrac{{50}}{{13}}cm;CH = \dfrac{{288}}{{13}}cm
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{AB}}{{AC}} = \dfrac{5}{{12}}\\
\Leftrightarrow \dfrac{{AB}}{5} = \dfrac{{AC}}{{12}} = k \Leftrightarrow \left\{ \begin{array}{l}
AB = 5k\\
AC = 12k
\end{array} \right.\\
Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Leftrightarrow {\left( {5k} \right)^2} + {\left( {12k} \right)^2} = {26^2}\\
\Leftrightarrow {k^2} = 4\\
\Leftrightarrow k = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
AB = 10\left( {cm} \right)\\
AC = 24\left( {cm} \right)
\end{array} \right.\\
{S_{ABC}} = \dfrac{1}{2}.AH.BC = \dfrac{1}{2}.AB.AC\\
\Leftrightarrow AH = \dfrac{{10.24}}{{26}} = \dfrac{{120}}{{13}}\left( {cm} \right)\\
Theo\,Pytago:\\
A{H^2} + B{H^2} = A{B^2}\\
\Leftrightarrow BH = \sqrt {{{10}^2} - {{\left( {\dfrac{{120}}{{13}}} \right)}^2}} = \dfrac{{50}}{{13}}\left( {cm} \right)\\
\Leftrightarrow CH = BC - BH = \dfrac{{288}}{{13}}\left( {cm} \right)\\
Vậy\,AB = 10cm;AC = 24cm;AH = \dfrac{{120}}{{13}}cm;\\
BH = \dfrac{{50}}{{13}}cm;CH = \dfrac{{288}}{{13}}cm
\end{array}$
