\[Схематический\ рисунок.\]
\[Дано:\ \ \]
\[MD\bot AB;\]
\[AD = BD;\]
\[BC = 16\ см;\]
\[P_{\text{AMC}} = 26\ см.\]
\[Найти:\]
\[\text{AC.}\]
\[Решение.\]
\[1)\ \mathrm{\Delta}AMD = \mathrm{\Delta}BMD - по\ первому\ \]
\[признаку:\]
\[\angle ADM = \angle BDM = 90{^\circ};\]
\[MD - общая\ сторона.\]
\[Отсюда:\]
\[AM = BM.\]
\[2)\ В\ \mathrm{\Delta}AMC:\]
\[P_{\text{AMC}} = AM + MC + AC = 26\]
\[BM + MC + AC = 26\]
\[BC + AC = 26\]
\[16 + AC = 26\]
\[AC = 10\ см.\]
\[Ответ:\ \ 10\ см.\]