\[\boxed{\mathbf{74.}еуроки - ответы\ на\ пятёрку}\]
\[Схематический\ рисунок.\]
\[Дано:\]
\[FM \parallel AB;\]
\[EM \parallel BC;\]
\[EF \parallel AC;\]
\[P_{\text{EBCA}} + P_{\text{BFCA}} + P_{\text{ABCM}} = 100\ см.\]
\[Найти:\]
\[P_{\text{ABC}}.\]
\[Решение.\]
\[1)\ Рассмотрим\ \]
\[параллелограмм\ EBCA:\]
\[AC = EB;\ \ \ BC = EA.\]
\[P_{\text{EBCA}} = EB + BC + CA + EA =\]
\[= AC + EA + AC + EA =\]
\[= 2AC + 2EA.\]
\[2)\ Рассмотрим\ \]
\[параллелограмм\ BFCA:\]
\[AB = CF;\ \ \ AC = BF.\]
\[P_{\text{BFCA}} = BF + FC + CA + AB =\]
\[= AC + AB + AC + AB =\]
\[= 2AC + 2AB.\]
\[3)\ Рассмотрим\ \]
\[параллелограмм\ ABCM:\]
\[AB = CM;\ \ \ BC = AM.\]
\[P_{\text{ABCM}} = AB + BC + CM + AM =\]
\[= AB + BC + AB + BC =\]
\[= 2AB + 2BC.\]
\[4)\ Треугольнике\ ABC:\]
\[P_{\text{ABC}} = AB + BC + CA\]
\[4P_{\text{ABC}} = 4AB + 4BC + 4CA\]
\[4P_{\text{ABC}} = P_{\text{EBCA}} + P_{\text{BFCA}} + P_{\text{ABCM}}\]
\[4P_{\text{ABC}} = 100\]
\[P_{\text{ABC}} = 25.\]
\[Ответ:\ \ 25\ см.\]