\[\boxed{\mathbf{782.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[AE = ED;\ \ \]
\[BE = GC.\]
\[\overrightarrow{\text{DC}} = \overrightarrow{a};\ \ \]
\[\overrightarrow{\text{BC}} = \overrightarrow{b}.\]
\[Выразить:\]
\[\overrightarrow{\text{EC}}\ \ \ и\ \ \ \overrightarrow{\text{AG}}.\]
\[\mathbf{Решение.}\]
\[1)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{AG}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BG}} = \overrightarrow{\text{DC}} + \overrightarrow{\frac{\text{BC}}{2}} =\]
\[= \overrightarrow{a} + \frac{1}{2}\overrightarrow{b}.\]
\[2)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{EC}} = \overrightarrow{\text{ED}} + \overrightarrow{\text{DC}} = \overrightarrow{\frac{\text{AD}}{2}} + \overrightarrow{\text{DC}} =\]
\[= \overrightarrow{\frac{\text{BC}}{2}} + \overrightarrow{\text{DC}} = \frac{1}{2}\overrightarrow{b} + \overrightarrow{a}.\]