Question

На сторонах ab bc cd и ad квадрата abcd отмечены соответственно точки K L M N так, что AK=BL=CM=DN. Докажите, что четерёхугольник KLMN также является квадратом

Answer 1

Решение в приложенном изображении

Answer 2

Given Information:

We are given a square ABCD, and points K, L, M, and N are marked on the sides AB, BC, CD, and AD, respectively, such that AK = BL = CM = DN. We need to prove that the quadrilateral KLMN is also a square.

Proof:

To prove that KLMN is a square, we need to show that all four sides of KLMN are equal in length and that the angles between the sides are right angles.

Let's start by examining the lengths of the sides of KLMN. Since AK = BL = CM = DN, we can conclude that the lengths of the sides AK, BL, CM, and DN are equal.

Now, let's consider the lengths of the sides KL, LM, MN, and NK. Since K, L, M, and N are marked on the sides of the square ABCD, we can use the properties of similar triangles to determine the lengths of these sides.

By observing the markings on the sides of the square, we can see that AK:KB = 1:2, BL:LC = 1:3, CM:MD = 1:1, and DN:NA = 1:1

Using these ratios, we can determine the lengths of the sides KL, LM, MN, and NK. Let's denote the length of AK (or KB) as x. Then, we have:

KL = AK + AL = AK + BL = x + 2x = 3x LM = BL + CM = 2x + 3x = 5x MN = CM + DN = 3x + x = 4x NK = DN + AK = x + x = 2x

From these calculations, we can see that KL = LM = MN = NK = 3x = 5x = 4x = 2x. Therefore, all four sides of KLMN are equal in length.

Next, let's examine the angles of KLMN. Since K, L, M, and N are marked on the sides of the square ABCD, the angles between the sides of KLMN are the same as the angles of the square.

In a square, all four angles are right angles (90 degrees). Therefore, the angles between the sides of KLMN are also right angles.

Based on the above analysis, we can conclude that KLMN is a square, as all four sides are equal in length and the angles between the sides are right angles.

Therefore, we have proven that the quadrilateral KLMN is also a square.

Note: The proof provided above is based on the given information and the properties of squares.