\[\boxed{\mathbf{804.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[AD = 3BC;\]
\[AK = \frac{1}{3}AD;\]
\[\overrightarrow{a} = \overrightarrow{\text{BA}};\ \ \]
\[\overrightarrow{b} = \overrightarrow{\text{CD}}.\]
\[\mathbf{Найти:}\]
\[\overrightarrow{\text{CK}} - \overrightarrow{\text{KD}}\ \ и\ \overrightarrow{\text{BC}}\ через\ \overrightarrow{a}\ и\ \overrightarrow{b}.\]
\[\mathbf{Решение.}\]
\[1)\ BC = \frac{\text{AD}}{3}\ \ и\ \ AK = \frac{1}{3}AD:\]
\[BC = AK.\]
\[2)\ BC \parallel AD;\ \ \]
\[AK \in AD\ (по\ условию):\ \ \]
\[AK \parallel BC.\]
\[3)\ По\ пункту\ 2:AK \parallel BC.\]
\[По\ пункту\ 1:\ \ BC = AK.\]
\[ABCK - параллелограмм\ \]
\[(по\ признаку).\]
\[Отсюда:\]
\[CK = AB;\ \text{\ \ }\]
\[\overrightarrow{\text{CK}} = \overrightarrow{\text{BA}} = \overrightarrow{a}.\text{\ \ }\]
\[4)\ \overrightarrow{\text{KD}} = \overrightarrow{\text{CD}} - \overrightarrow{\text{CK}} = \overrightarrow{b} - \overrightarrow{a}.\]
\[5)\ \overrightarrow{\text{BC}} = \frac{1}{2}\overrightarrow{\text{KD}} = \frac{1}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a}.\]
\[Ответ:\ \overrightarrow{\text{CK}} = \overrightarrow{a};\ \ \overrightarrow{\text{KD}} = \overrightarrow{b} - \overrightarrow{a};\ \]
\[\overrightarrow{\text{BC}} = \frac{1}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a}.\]