Решебник по геометрии 8 класс Мерзляк Задание 412

\[Схематический\ рисунок.\]

\[Дано:\]

\[BK - медиана;\]

\[AM\ :MB = 4\ :3.\]

\[Найти:\]

\[1)\ OM\ :CO;\]

\[2)\ KO\ :BO.\]

\[Решение.\]

\[1)\ Проведем\ прямую:\]

\[ME \parallel BK;\ \ \ \]

\[ME \cap AC = E.\]

\[В\ \mathrm{\Delta}ABK:\]

\[ME \parallel BK;\ \ \ \]

\[\frac{\text{AM}}{\text{AE}} = \frac{\text{BM}}{\text{KE}}\]

\[\frac{\text{AE}}{\text{KE}} = \frac{\text{AM}}{\text{BM}} = \frac{4}{3}\text{\ \ \ }\]

\[AE = \frac{4}{3}\text{KE.}\]

\[В\ \mathrm{\Delta}ABC:\]

\[CK = AK = \frac{1}{2}AC;\]

\[AK = KE + AE = \frac{7}{3}KE;\]

\[KE = \frac{3}{7}AK = \frac{3}{14}\text{AC.}\]

\[В\ \mathrm{\Delta}MCE:\]

\[OK \parallel ME;\ \ \ \]

\[\frac{\text{OM}}{\text{KE}} = \frac{\text{CO}}{\text{CK}}\]

\[\frac{\text{OM}}{\text{CO}} = \frac{\text{KE}}{\text{CK}} = \frac{3}{7}.\]

\[2)\ Проведем\ прямую:\]

\[KD \parallel CM;\ \ \]

\[KD \cap AB = D.\]

\[В\ \mathrm{\Delta}AMC:\]

\[CK = AK;\ \ \ \]

\[KD \parallel CM;\]

\[MD = AD.\]

\[В\ \mathrm{\Delta}ABC:\]

\[AM = MD + AD = 2MD\]

\[AM = \frac{4}{3}\text{MB\ \ \ }\]

\[MD = \frac{2}{3}\text{MB.}\]

\[В\ \mathrm{\Delta}BKD:\]

\[OM \parallel KD;\ \ \ \]

\[\frac{\text{BO}}{\text{BM}} = \frac{\text{KO}}{\text{MD}}\]

\[\frac{\text{KO}}{\text{BO}} = \frac{\text{MD}}{\text{BM}} = \frac{2}{3}.\]

\[Ответ:\ \ 1)\ 3\ :7;\ 2)\ 2\ :3.\]