Решебник по геометрии 7 класс Мерзляк Задание 281

\[\mathbf{Схематический\ рисунок.}\]

\[Дано:\ \ \]

\[AM - медиана\ \mathrm{\Delta}ABC;\]

\[A_{1}M_{1} - медиана\ \mathrm{\Delta}A_{1}B_{1}C_{1};\]

\[AM = A_{1}M_{1};\]

\[\angle BAM = \angle B_{1}A_{1}M_{1};\]

\[\angle MAC = \angle M_{1}A_{1}C_{1}.\]

\[Доказать:\]

\[\mathrm{\Delta}ABC = \mathrm{\Delta}A_{1}B_{1}C_{1}.\]

\[\mathbf{Доказательство.}\]

\[1)\ Отложим\ отрезки\ на\ \]

\[прямых\ \text{AM\ }и\ A_{1}M_{1}\ :\]

\[MD = AM;\ \ \ \]

\[M_{1}D_{1} = A_{1}M_{1}.\]

\[2)\ \mathrm{\Delta}AMC = \mathrm{\Delta}DMB - по\ первому\ \]

\[признаку:\]

\[MC = MB = \frac{1}{2}BC;\]

\[\angle AMC = \angle DMB - вертикальные.\]

\[Отсюда:\]

\[AC = BD;\ \ \ \]

\[\angle MAC = \angle MDB.\]

\[3)\ Аналогично:\]

\[A_{1}C_{1} = B_{1}D_{1};\ \ \ \]

\[\angle M_{1}A_{1}C_{1} = \angle M_{1}D_{1}B_{1}.\]

\[4)\ \mathrm{\Delta}ABD = \mathrm{\Delta}A_{1}B_{1}D_{1} - по\ второму\ \]

\[признаку:\]

\[AD = 2AM = 2A_{1}M_{1} = A_{1}D_{1};\]

\[\angle B_{1}D_{1}A_{1} = \angle B_{1}A_{1}D_{1} =\]

\[= \angle BAD = \angle BDA.\]

\[Отсюда:\]

\[AB = A_{1}B_{1};\ \ \ \]

\[BD = B_{1}D_{1}.\]

\[5)\ \mathrm{\Delta}ABC = \mathrm{\Delta}A_{1}B_{1}C_{1} - по\ первому\ \]

\[признаку:\]

\[AC = BD = B_{1}D_{1} = A_{1}C_{1};\ \ \ \]

\[\angle BAC = \angle B_{1}A_{1}C_{1}.\]

\[Что\ и\ требовалось\ доказать.\]