Đáp án:
$\begin{array}{l}
a)\dfrac{{AB}}{{AC}} = \dfrac{5}{6} \Leftrightarrow AB = \dfrac{5}{6}AC\\
Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Leftrightarrow {\left( {\dfrac{5}{6}AC} \right)^2} + A{C^2} = {122^2}\\
\Leftrightarrow A{C^2} = 8784\\
\Leftrightarrow AC = 12\sqrt {61} \\
\Leftrightarrow AB = \dfrac{5}{6}AC = 10\sqrt {61} \\
Do:A{B^2} = BH.BC\\
\Leftrightarrow BH = \dfrac{{{{\left( {10\sqrt {61} } \right)}^2}}}{{122}} = 50\\
\Leftrightarrow CH = 122 - 50 = 72\\
Vay\,BH = 50cm;CH = 72cm\\
b)Theo\,t/c:\\
\dfrac{{AD}}{{AB}} = \dfrac{{CD}}{{BC}}\\
\Leftrightarrow \dfrac{{AD}}{{10\sqrt {61} }} = \dfrac{{CD}}{{122}} = \dfrac{{AD + CD}}{{10\sqrt {61} + 122}} = \dfrac{{12\sqrt {61} }}{{10\sqrt {61} + 122}}\\
\Leftrightarrow AD = 36,58\left( {cm} \right)
\end{array}$
