Question

Дан шестиугольник A1A2A3A4A5A6. Его стороны А1А2 и А4А5, А2А3 и А5А6, А3А4 и А6А1, попарно равны и параллельны. Используя центральную симметрию, докажите, что диагонали А1А4 и А2А5, А3А6 данного шестиугольника пересекаются в одной точке. Блин срочняк надо, помогите плиз!!!

Answer 1

Треугольники, образованные парными боковыми сторонами и парой из указанных диагоналей, равны. Например, треугольник А1А2О = треугольник А4А5О, где О - точка пересечения А1А4 и А2А5 => обе эти диагонали в точке их пересечения делятся пополам. И эта пара сторон и пара диагоналей центрально симметрична относительно О. Аналогично и для другой пары сторон. Мы видим, что и они делятся точкой пересечения пополам, то есть эта точка совпадает с О. Поэтому у фигуры есть центр симметрии, и все диагонали, соединяющие центрально симметричные вершины (А1 и А4, А2 и А5, А4 и А6), обязательно проходят через центр симметрии и делятся им пополам.  

Answer 2

Proof that the diagonals of the given hexagon intersect at one point using central symmetry

To prove that the diagonals A1A4 and A2A5, A3A6 of the given hexagon intersect at one point, we can use the concept of central symmetry.

Let's consider the hexagon A1A2A3A4A5A6, where the sides A1A2 and A4A5, A2A3 and A5A6, A3A4 and A6A1 are pairwise equal and parallel.

To begin the proof, we will use the concept of central symmetry. Central symmetry is a transformation that maps each point P to its symmetric point P' with respect to a center O. In this case, we will consider the center of symmetry to be the intersection point of the diagonals A1A4 and A2A5, which we need to prove.

Now, let's consider the hexagon A1'A2'A3'A4'A5'A6', which is the image of the original hexagon A1A2A3A4A5A6 under central symmetry with respect to the center O.

Since the sides A1A2 and A4A5, A2A3 and A5A6, A3A4 and A6A1 are pairwise equal and parallel in the original hexagon, their corresponding sides in the image hexagon A1'A2'A3'A4'A5'A6' will also be equal and parallel.

Therefore, we can conclude that the image hexagon A1'A2'A3'A4'A5'A6' is also a regular hexagon.

Now, let's consider the diagonals of the image hexagon A1'A2'A3'A4'A5'A6'. The diagonals A1'A4' and A2'A5', A3'A6' of the image hexagon will also be equal and parallel to each other, just like the diagonals A1A4 and A2A5, A3A6 of the original hexagon.

Since the diagonals A1'A4' and A2'A5', A3'A6' of the image hexagon are equal and parallel, they must intersect at one point. Let's denote this intersection point as P'.

Now, since the image hexagon A1'A2'A3'A4'A5'A6' is obtained by applying central symmetry to the original hexagon A1A2A3A4A5A6, the diagonals A1A4 and A2A5, A3A6 of the original hexagon must also intersect at the corresponding point P.

Therefore, we have proved that the diagonals A1A4 and A2A5, A3A6 of the given hexagon intersect at one point.

Note: The proof above is based on the concept of central symmetry. If you need further clarification or have any additional questions, please let me know.