\[\boxed{\mathbf{764.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ \left( \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} - \overrightarrow{\text{MC}} \right) + \left( \overrightarrow{\text{MD}} - \overrightarrow{\text{KD}} \right) =\]
\[= \left( \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CM}} \right) + \left( \overrightarrow{\text{MD}} + \overrightarrow{\text{DK}} \right) =\]
\[= \overrightarrow{\text{AM}} + \overrightarrow{\text{MK}} = \overrightarrow{\text{AK}}.\]
\[\textbf{б)}\ \left( \overrightarrow{\text{CB}} + \overrightarrow{\text{AC}} + \overrightarrow{\text{BD}} \right) - \left( \overrightarrow{\text{MK}} + \overrightarrow{\text{KD}} \right) =\]
\[= \left( \overrightarrow{\text{AC}} + \overrightarrow{\text{CB}} + \overrightarrow{\text{BD}} \right) - \overrightarrow{\text{MD}} =\]
\[= \overrightarrow{\text{AD}} + \overrightarrow{\text{DM}} = \overrightarrow{\text{AM}}.\]
\[\boxed{\mathbf{764.еуроки - ответы\ на\ пятёрку}}\]
\[1)\ Рисунок\ по\ условию\ задачи:\]
\[\mathbf{а)}\]
\(\mathbf{\ }\)
\[\mathbf{б)\ }\]
\[\mathbf{Дано:}\]
\[окружность\ (O,\ R);\ \]
\[AB = CD;\]
\[\angle AOB = 112{^\circ}\]
\[\mathbf{Доказать:}\]
\[\cup AB = \cup CD.\]
\[Найти:\]
\[\cup CD - ?\ \]
\[\cup CBD - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}ABO = \mathrm{\Delta}OCD - по\ трем\ \]
\[сторонам:\]
\[BO = AO = OC = OD = r;\]
\[AB = CD\ (по\ условию).\]
\[Отсюда:\]
\[2)\ \angle BOA = \cup AB\ \]
\[(как\ центральный);\ \]
\[\angle COD = \cup CD\ \]
\[(как\ центральный);\]
\[\ \angle BOA = \angle COD.\]
\[Следовательно:\ \]
\[\cup AB = \cup CD.\]
\[Что\ и\ требовалось\ доказать.\]
\[3)\ \angle BOA = \angle COD = 112{^\circ}\ \]
\[(как\ вертикальные):\]
\[\angle COD = \cup CD = 112{^\circ}\ \]
\[(как\ центральный).\]
\[4)\ \cup CBD = 360{^\circ} - 112{^\circ} =\]
\[= 248{^\circ}.\]
\[Ответ:\ \cup CD = 112{^\circ}\ и\ \]
\(\cup CBD = 248{^\circ}.\)