[tex]\it \Delta ABC-dr,\ \hat A=90^o,\ \hat B=30^o\ \stackrel{T.\angle 30^o}{\Longrightarrow}\ AC=\dfrac{BC}{2}=\dfrac{72}{2}=36\ cm\\ \\ \\ \Delta ABC-dr,\ \hat A=90^o,\ \stackrel{T. P.}{\Longrightarrow}\ AB^2=BC^2-AC^2=(BC-AC)(BC+AC)=\\ \\ = (72-36)(72+36)=36\cdot108=36\cdot36\cdot3=36^2\cdot3 \Rightarrow AB=36\sqrt3\ cm[/tex]
[tex]\it AD=\dfrac{AB\cdot AC}{BC}=\dfrac{\ 36\sqrt3\cdot36^{(36}}{72}=\dfrac{36\sqrt3}{2}=18\sqrt3\ cm[/tex]
