\[\boxed{\mathbf{851.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[ABCD - выпуклый\ \]
\[четырехугольник;\]
\[BE = EF = FC;\]
\[AG = GH = HD.\]
\[\mathbf{Доказать:}\]
\[S_{\text{EFHG}} = \frac{1}{3}S_{\text{ABCD}}.\]
\[\mathbf{Доказательство.}\]
\[Допустим:\]
\[AG = GH = HD = a;\]
\[BE = EF = FC = b.\]
\[2)\ BE = EF;\ B_{1}E_{1} = E_{1}F_{1} = d_{1}:\]
\[DB_{1} + DE_{1} + DF_{1} =\]
\[= \left( DE_{1} - d_{1} \right) + DE_{1} + \left( DE_{1} + d_{1} \right) =\]
\[= 3DE_{1}.\]
\[3)\ Аналогично:\]
\[KG_{1} + KH_{1} + KD_{1} = 3KH_{1}.\]
\[4)\ S_{\text{ABCD}} =\]
\[= \frac{3}{2} \bullet \left( a \bullet DE_{1} + b \bullet KH_{1} \right).\]
\[5)\ S_{\text{EFBG}} = S_{\text{GEH}} + S_{\text{EFH}} =\]
\[= \frac{a}{2}DE_{1} + \frac{b}{2}KH_{1} =\]
\[= \frac{1}{2} \bullet \left( a \bullet DE_{1} + b \bullet KH_{1} \right).\]
\[6)\ Получаем:\]
\[\frac{S_{\text{EFHG}}}{S_{\text{ABCD}}} = \frac{\frac{1}{2} \bullet \left( a \bullet DE_{1} + b \bullet KH_{1} \right)}{\frac{3}{2} \bullet \left( a \bullet DE_{1} + b \bullet KH_{1} \right)} =\]
\[= \frac{1}{3}.\]
\[S_{\text{EFHG}} = \frac{1}{3}S_{\text{ABCD}}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]