\[\boxed{\mathbf{812.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[a_{1} - a_{4} = a_{5} - a_{2} = a_{3} - a_{6};\]
\[A_{1}A_{2} = a_{1};A_{2}A_{3} = a_{2};\]
\[A_{3}A_{4} = a_{3};A_{4}A_{5} = a_{4};\]
\[A_{5}A_{6} = a_{5};A_{6}A_{1} = a_{6}.\]
\[\mathbf{Доказать:}\]
\[\angle A_{1} = \angle A_{2} = \ldots = \angle A_{6};\]
\[A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} - выпуклый\ \]
\[шестиугольник.\]
\[\mathbf{Доказательство.}\]
\[a_{1} - a_{4} + a_{2} + a_{3} + a_{4} =\]
\[= a_{5} - a_{2} + a_{2} + a_{3} + a_{4} =\]
\[= a_{3} - a_{6} + a_{2} + a_{3} + a_{4}\]
\[a_{5} - a_{2} = a_{3} - a_{6} \Longleftrightarrow a_{2} =\]
\[= a_{5} + a_{6} - a_{3}.\]
\[Получаем:\]
\[a_{1} - a_{4} = a_{3} - a_{6}\ \ \ \ или\ \ \ \]
\[a_{4} = a_{1} - a_{3} + a_{6}.\]
\[2)\ a_{3} - a_{6} + a_{2} + a_{3} + a_{4} =\]
\[3)\ a_{1} + a_{2} + a_{3} = a_{5} + a_{3} + a_{4} =\]
\[= a_{5} + a_{6} + a_{1}.\]
\[4)\ AB = a_{1} + a_{2} + a_{3};\]
\[BC = a_{5} + a_{3} + a_{4};\]
\[AC = a_{5} + a_{6} + a_{1}:\]
\[\ \mathrm{\Delta}ABC - равносторонний.\ \]
\[5)\ AA_{2} = AA_{1} = a_{1};\ \ \]
\[A_{3}B = BA_{4} = a_{3};\ \]
\[A_{5}C = CA_{6} = a_{5}\]
\[и\ \]
\[\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 5 =\]
\[= \angle 6 = \frac{180{^\circ} - 60{^\circ}}{2} = 60{^\circ};\]
\[то\ равнобедренные\ \]
\[треугольники\]
\[\text{\ A}A_{2}A_{1};A_{3}BA_{4};\]
\[A_{5}CA_{6} - равносторонние.\]
\[6)\ Так\ как\ все\ углы\ \]
\[шестиугольника\ смежные\ \]
\[с\ \angle 60{^\circ}:\]
\[\angle A_{1} = \angle A_{2} = \angle A_{3} = \angle A_{4} =\]
\[= \angle A_{5} = \angle A_{6} = 180 - 60 =\]
\[= 120{^\circ}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]