Question

Докажите, что середины сторон прямоугольника являются вершинами ромба.

Answer 1

 

Пусть четырёхугольник ABCD. Пусть M, N, K, L соотв. середины его сторон AB, BC, CD и AD. Тогда в треугольнике ABC: MN является средней линией, значит, равна половине диагонали BC четырёхугольника. Аналогично доказываем, что NK=1/2 AC, KL=1/2 BC, LM=1/2 AC. Но так как AC=BC получаем, что MN=NK=KL=LM.

Answer 2

Proving that the Midpoints of the Sides of a Rectangle Form a Rhombus

To prove that the midpoints of the sides of a rectangle form a rhombus, we can use the properties of rectangles and rhombuses.

1. Properties of a Rectangle: - A rectangle is a quadrilateral with four right angles. - Opposite sides of a rectangle are equal in length.

2. Properties of a Rhombus: - A rhombus is a quadrilateral with all sides of equal length. - The diagonals of a rhombus are perpendicular bisectors of each other.

Proof:

Let's consider a rectangle ABCD with midpoints E, F, G, and H on sides AB, BC, CD, and DA, respectively.

Claim: The midpoints of the sides of a rectangle form a rhombus.

Proof: 1. Equal Opposite Sides: - Since ABCD is a rectangle, AB = CD and BC = AD. - By definition of midpoints, E is the midpoint of AB, F is the midpoint of BC, G is the midpoint of CD, and H is the midpoint of DA. - Therefore, AE = EB, BF = FC, CG = GD, and DH = HA.

2. Diagonals Bisect Each Other: - The diagonals of the rectangle ABCD are AC and BD. - The midpoints of the diagonals AC and BD are E and G, and F and H, respectively. - Since E, F, G, and H are midpoints, EF = GH, and EG = FH.

3. Conclusion: - From the above properties, it is evident that EFGH is a rhombus, as it has all sides of equal length and its diagonals bisect each other.

Therefore, the midpoints of the sides of a rectangle form a rhombus.

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