\[Схематический\ рисунок.\]
\[Дано:\ \ \]
\[BD - медиана;\]
\[P_{\text{ABC}} = 60\ см;\]
\[P_{\text{ABD}} = 36\ см;\]
\[P_{\text{CBD}} = 50\ см.\]
\[Найти:\]
\[\text{BD.}\]
\[Решение.\]
\[1)\ В\ \ \mathrm{\Delta}ABD:\]
\[P_{\text{ABD}} = AB + BD + AD = 36\]
\[AB + AD = 36 - BD\]
\[2)\ В\ \mathrm{\Delta}CBD:\]
\[P_{\text{CBD}} = CB + BD + CD = 50\]
\[CB + CD = 50 - BD.\]
\[3)\ В\ \mathrm{\Delta}ABC:\]
\[P_{\text{ABC}} = AB + BC + AC = 60\]
\[AB + BC + AD + CD = 60\]
\[36 - BD + 50 - BD = 60\]
\[2BD = 26\ \ \ \]
\[BD = 13\ см.\]
\[Ответ:\ \ 13\ см.\]