Question

Ребята, помогите, кто сможет! Решить легенькую задачку. :)Вот, собственно, и сама задачка: Сторона AB и дмагональ BD прямоугольника ABCD лежат в плоскости альфа. Докажите, что и вершина C прямоугольника лежит в той же плоскости.

Answer 1

Может быть так:
АВ и ВД пересекаются, поэтом улежат в одной плоскости. ВД и ВС пересекаются, значит тоже лежат в одной плоскости, но ВС пересекается и с ВА, они тоже лежат в одной плоскости. Отсюда следует, что и АВ и ВД и ВС лежат в одной плоскости. А раз прямая ВС лежит в плоскости, то и все точки этой прямой лежает в этой плоскости. Соответственно, точка С лежит в плоскости альфа. 

Answer 2

Proof that vertex C of rectangle ABCD lies in the same plane as sides AB and diagonal BD:

To prove that vertex C of rectangle ABCD lies in the same plane as sides AB and diagonal BD, we can use the fact that any three non-collinear points determine a unique plane in three-dimensional space.

Let's consider points A, B, and D. Since AB and BD are both in the plane alpha, points A, B, and D are collinear in that plane. Now, we need to show that point C also lies in the same plane.

To do this, we can use the fact that a rectangle is a quadrilateral with four right angles. Since ABCD is a rectangle, angle B is a right angle. This means that line BC is perpendicular to line AB.

Now, let's consider the plane beta that contains points A, B, and D. Since AB and BD are both in the plane alpha, and line BC is perpendicular to line AB, we can conclude that line BC is also perpendicular to plane alpha. Therefore, line BC is contained entirely within plane beta.

Since line BC is contained entirely within plane beta, and points A, B, and D are collinear in plane alpha, we can conclude that point C lies in both plane alpha and plane beta. Therefore, vertex C of rectangle ABCD lies in the same plane as sides AB and diagonal BD.

In conclusion, we have proven that vertex C of rectangle ABCD lies in the same plane as sides AB and diagonal BD.