Question

Если три вершины прямоугольника лежат на окружности, то докажите, что и четвёртая вершина тоже лежит на этой окружности

Answer 1

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Answer 2

Proof that the Fourth Vertex of a Rectangle Lies on the Circumference

To prove that the fourth vertex of a rectangle lies on the circumference when three vertices lie on the circumference, we can use the properties of rectangles and circles.

Let's assume that we have a rectangle ABCD, where A, B, and C are three vertices that lie on the circumference of a circle. We want to prove that the fourth vertex, D, also lies on the same circumference.

To begin the proof, we need to establish some properties of rectangles and circles:

1. In a rectangle, opposite sides are equal in length. 2. In a rectangle, diagonals are equal in length and bisect each other. 3. In a circle, any two points on the circumference are equidistant from the center.

Now, let's proceed with the proof:

1. Since ABCD is a rectangle, we know that opposite sides are equal in length. Therefore, AB = CD and BC = AD. 2. The diagonals of a rectangle are equal in length and bisect each other. Therefore, AC = BD. 3. Since A, B, and C lie on the circumference of a circle, they are equidistant from the center of the circle. 4. Let O be the center of the circle. Since A, B, and C are equidistant from O, we have OA = OB = OC. 5. By the properties of rectangles, we know that AC = BD. Therefore, AC = BD = 2 * OA. 6. Since AC = BD = 2 * OA and OA = OB = OC, we can conclude that AC = BD = 2 * OB. 7. From step 6, we can infer that D is equidistant from O, just like A, B, and C. Therefore, D also lies on the circumference of the circle.

Thus, we have proven that if three vertices of a rectangle lie on the circumference of a circle, the fourth vertex also lies on the same circumference.

Please note that this proof assumes that the given figure is a rectangle and that the three vertices lie on the circumference of a circle. If any of these assumptions are not met, the conclusion may not hold.