\[\boxed{\mathbf{749.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\]
\[ABCD - равнобедренная\ \]
\[трапеция;\]
\[\text{S\ }и\ T - середины\ боковых\ \]
\[сторон.\]
\[Равны\ ли\ векторы\]
\[\textbf{а)}\ \overrightarrow{\text{NL}}\ и\ \overrightarrow{\text{KL}}:\]
\[\overrightarrow{\text{NL}}\ и\ \overrightarrow{\text{KL}} - не\ коллинеарны;\]
\[\overrightarrow{\text{NL}} \neq \overrightarrow{\text{KL}}.\]
\[\textbf{б)}\ \overrightarrow{\text{MS}}\ и\ \overrightarrow{\text{SN}}:\]
\[\overrightarrow{\text{MS}} \uparrow \uparrow \overrightarrow{\text{SN}}\ и\ \left| \overrightarrow{\text{MS}} \right| =\]
\[= \left| \overrightarrow{\text{SN}} \right|\ \left( так\ как\ S - середина\ \text{MN} \right);\]
\[\ \overrightarrow{\text{MS}} = \overrightarrow{\text{SN}}.\]
\[\textbf{в)}\ \overrightarrow{\text{MN}}\ и\ \overrightarrow{\text{KL}}:\]
\[\overrightarrow{\text{MN}}\ и\ \overrightarrow{\text{KL}} - не\ коллинеарны;\]
\[\ \overrightarrow{\text{MN}} \neq \overrightarrow{\text{KL}}.\]
\[\textbf{г)}\ \overrightarrow{\text{TS}}\ и\ \overrightarrow{\text{KM}}:\]
\[TS = \frac{1}{2}\text{KM\ }\]
\[(по\ свойству\ средней\ линии);\]
\[\overrightarrow{\text{TS}} \uparrow \uparrow \overrightarrow{\text{KM}}\ и\ \left| \overrightarrow{\text{TS}} \right| \neq \left| \overrightarrow{\text{KM}} \right|;\]
\[\ \overrightarrow{\text{TS}} \neq \overrightarrow{\text{KM}}.\]
\[\textbf{д)}\ \overrightarrow{\text{TL}}\ и\ \overrightarrow{\text{KT}}:\]
\[\overrightarrow{\text{TL}} \uparrow \uparrow \overrightarrow{\text{KT}}\ и\ \left| \overrightarrow{\text{TL}} \right| =\]
\[= \left| \overrightarrow{\text{KT}} \right|\ \left( так\ как\ T - середина\ \text{KL} \right);\ \]
\[\overrightarrow{\text{TL}} = \overrightarrow{\text{KT}}.\]
\[Ответ:а)\ нет;б)\ да;в)\ нет;\]
\[\textbf{г)}\ нет;д)\ да.\]