[tex]\displaystyle\\
\text{Se da:}\\
\Delta ABC~\text{isoscel cu}~AB=AC=8~cm,~BC = 8\sqrt{3}~cm\\\\
\text{Se cere:}\\
\text{Inaltimea }~AD\perp BC,~D\in BC\\\\
\text{Rezolvare:}\\
\text{In triunghiul isoscel, {\bf inaltimea} dusa din varful unghiului format de}\\
\text{laturile congruente, cate pe mijlocul laturii opuse.}\\\\
\Longrightarrow~BD=CD=\frac{BC}{2}=\frac{8\sqrt3}{2}=4\sqrt3~cm\\\\
[/tex]
[tex]\displaystyle\\
\text{Alegem triunghiul dreptunghic }\Delta ABD~\text{ in care avem:}\\
AB=8~cm=ipotenuza\\
BD=4\sqrt3=cateta\\
AD=? = cateta\\\\
AD=\sqrt{AB^2-BD^2}= \sqrt{8^2-\Big(4\sqrt{3}\Big)^2}= \sqrt{64-48} =\sqrt{16}=\boxed{4~cm}
[/tex]
