\[\boxed{\mathbf{615.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[BD \cap AC = O;\]
\[EF \parallel AD;\]
\[BC = b;\]
\[AD = a.\]
\[\mathbf{Найти:}\]
\[EF - ?\]
\[\mathbf{Решение.}\]
\[1)\ \ \mathrm{\Delta}BOC\sim\mathrm{\Delta}AOD - по\ двум\ \]
\[углам:\]
\[\angle BOC = \angle AOD\ \]
\[(как\ вертикальные);\]
\[\ \angle BCA = \angle CAD\ \]
\[(как\ накрестлежащие).\ \]
\[Отсюда:\]
\[\frac{\text{AD}}{\text{BC}} = \frac{\text{OD}}{\text{BO}} = \frac{\text{AO}}{\text{OC}} = \frac{a}{b}.\]
\[2)\ \mathrm{\Delta}ABD\sim\mathrm{\Delta}EBO\ \]
\[(по\ двум\ углам):\]
\[\angle ABD - общий\ ;\]
\[\angle BEO = \angle BAD\ \]
\[(как\ соответственные).\]
\[Отсюда:\]
\[\frac{\text{BD}}{\text{BO}} = \frac{\text{AB}}{\text{BE}} = \frac{\text{AD}}{\text{EO}}\]
\[\frac{\text{AD}}{\text{EO}} = \frac{\text{BD}}{\text{BO}}\]
\[EO = \frac{AD \bullet BO}{\text{BD}} = \frac{a \bullet BO}{\text{BD}}.\]
\[3)\ \mathrm{\Delta}BDC\sim\mathrm{\Delta}ODF\ \]
\[(по\ двум\ углам):\]
\[\angle CDB - общий;\]
\[\angle BCD = \angle OFD\ \]
\[(как\ соответственные).\]
\[Отсюда:\]
\[\frac{\text{BD}}{\text{OD}} = \frac{\text{CD}}{\text{DF}} = \frac{\text{BC}}{\text{OF}}\]
\[\frac{\text{BC}}{\text{OF}} = \frac{\text{BD}}{\text{OD}}\]
\[OF = \frac{BC \bullet OD}{\text{BD}} = \frac{b \bullet OD}{\text{BD}}.\]
\[4)\ EF = EO + OF =\]
\[= \frac{AD \bullet BO}{\text{BD}} + \frac{BC \bullet OD}{\text{BD}} =\]
\[= \frac{a \bullet BO}{\text{BD}} + \frac{b \bullet OD}{\text{BD}} =\]
\[= \frac{a \bullet BO + b \bullet OD}{\text{BD}}.\]
\[5)\ BD = OB + OD =\]
\[= OB + \frac{a \bullet BO}{b} =\]
\[= \frac{b \bullet OB + a \bullet OB}{b} = \frac{\text{OB}(b + a)}{b}.\]
\[6)\ EF = \frac{a \bullet OB + b \bullet \frac{a \bullet OB}{b}}{\frac{\text{OB}(b + a)}{b}} =\]
\[= \frac{a \bullet OB + a \bullet OB}{\frac{\text{OB}(b + a)}{b}} =\]
\[= \frac{\text{OB}(a + a)}{\frac{\text{OB}(b + a)}{b}} =\]
\[= \frac{2a \bullet OB \bullet b}{\text{OB}(b + a)} = \frac{2ab}{b + a}.\]
\[\mathbf{Ответ:}EF = \frac{2ab}{b + a}\mathbf{.}\]
\[\boxed{\mathbf{615}\mathbf{.}\mathbf{еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\mathbf{\ }\mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[D \in \text{BC};\]
\[\text{ED} \parallel \text{AC};\ \]
\[\text{DF} \parallel \text{AB}.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ACD}} = S_{\text{BDF}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ FC \parallel ED\ и\ EF \parallel DC:\ \]
\[EDFC - параллелограмм.\]
\[2)\ EF \parallel BD\ и\ FD \parallel EB;\ \]
\[FDEB - параллелограмм.\]
\[3)\ ED - диагональ\ FDEB;\]
\[FD - диагональ\ EDFC:\]
\[S_{\text{EDF}} = \frac{1}{2}S_{\text{FDEB}}\ и\ S_{\text{EDF}} = \frac{1}{2}S_{\text{EDFC}}\]
\[\frac{1}{2}S_{\text{FDEB}} = \frac{1}{2}S_{\text{EDFC}}.\]
\[4)\ S_{\text{ACD}} = \frac{1}{2}S_{\text{EDFC}};\]
\[S_{\text{BDF}} = \frac{1}{2}S_{\text{FDEB}}.\]
\[5)\ S_{\text{ACD}} = S_{\text{BDF}}.\]
\[Что\ и\ требовалось\ доказать.\]