\[\boxed{\mathbf{848.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC\ \left( \text{AB} \neq \text{AC} \right);\]
\[M \in BC;BM = MC;\]
\[AF - биссектриса\ \angle A;\]
\[MD \parallel AF;D \in AB;\]
\[F = MD \cap AC.\]
\[\mathbf{Доказать:}\]
\[BD = CE.\]
\[\mathbf{Доказательство.}\]
\[1)\ \mathrm{\Delta}DAE - равнобедренный,\ с\ \]
\[основанием\ DE:\]
\[MD \parallel AF \Longrightarrow \angle ADE = \angle BAF =\]
\[= \angle\frac{A}{2};\]
\[\angle DAE = 180{^\circ} - \angle A\ \Longrightarrow \angle AED =\]
\[= 180 - \left( 180 - \angle A + \angle\frac{A}{2} \right) =\]
\[= \angle\frac{A}{2}.\]
\[Следовательно:AD = AE.\]
\[2)\ Пусть\ BM = MC = d;\ \ \]
\[FM = f:\]
\[MD \parallel AF;\ \ \angle B - общий;\ \ \]
\[\mathrm{\Delta}ABF\sim\mathrm{\Delta}DMB;\]
\[k_{1} = \frac{\text{BA}}{\text{BD}} = \frac{\text{BF}}{\text{BM}} = \frac{d - f}{d}\text{.\ }\]
\[Получаем:\]
\[AD = BD - BA = \left( 1 - \frac{d - f}{d} \right) =\]
\[= \frac{f}{d}\text{BD.}\]
\[3)\ MD \parallel AF;\ \ \]
\[\angle B - общий \Longrightarrow \mathrm{\Delta}MCE\sim\mathrm{\Delta}FCA;\]
\[k_{2} = \frac{\text{CE}}{\text{CA}} = \frac{\text{CM}}{\text{CF}} = \frac{d}{d + f}\text{.\ }\]
\[Отсюда:\]
\[AE = CA - CE =\]
\[= \left( \frac{d + f}{d} - 1 \right) \bullet CE = \frac{f}{d}\text{CE.}\]
\[4)Получаем:\]
\[AD = AE;\ \]
\[\frac{f}{d}BD = \frac{f}{d}CE;\ \ \]
\[BD = CE.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]
\[\boxed{\mathbf{848.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[\text{AD} \parallel BC;\ \ AD > BC;\]
\[E;F \in AD;\]
\[\text{BE} \parallel CF;\ \ \]
\[O = AC \cap BD;\]
\[G = AC \cap BE;\ \]
\[H = BD \cap CF.\]
\[\mathbf{Доказать:}\]
\[S_{\text{EGOHF}} = S_{\text{ABG}} + S_{\text{BCO}} + S_{\text{DCH}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ S_{\text{ABC}} = S_{\text{BCD}} = \frac{1}{2}BC \bullet h.\]
\[2)\ S_{\text{ABG}} + S_{\text{BGO}} + S_{\text{BOC}} =\]
\[= S_{\text{DCH}} + S_{\text{COH}} + S_{\text{BOC}}.\]
\[3)\ S_{\text{EBCF}} = BC \bullet h = 2S_{\text{ABC}} =\]
\[= S_{\text{ABC}} + S_{\text{BCD}}.\]
\[4)\ S_{\text{EGOHF}} + S_{\text{BGO}} + S_{\text{BOC}} + S_{\text{COH}} =\]
\[5)\ S_{\text{EGOHF}} = S_{\text{ABG}} + S_{\text{BCO}} + S_{\text{DCH}}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]